GSEB Class 12 Statistics Solutions Chapter 4 Limit

GSEB Solutions Class 12 Statistics Part 2 Chapter 4 Limit Ex 4

Gujarat Board Textbook Solutions Class 12 Statistics Part 2 Chapter 4 Limit Ex 4

Section A

 

Question 1. What is the modulus form of 0.3 neighbourhood of 3?
(a) \(|x - 0.3| < 3\)
(b) \(|x - 3| < 0.3\)
(c) \(|x + 3| < 0.3\)
(d) \(|x - 3| > 0.3\)
Answer: (b) \(|x - 3| < 0.3\)
In simple words: The modulus form for the 0.3 neighbourhood of 3 means all numbers 'x' that are within a distance of 0.3 from 3. This is written as \(|x - 3| < 0.3\).

🎯 Exam Tip: Understanding the definition of neighbourhood in modulus form is crucial for questions like this, which test basic concepts directly.

 

Question 2. What is the interval form of 0.02 neighbourhood of -2?
(a) (1.98, 2.02)
(b) (2, 0.02)
(c) (- 2.02, -1.98)
(d) (- 2.02, 1.98)
Answer: (c) (- 2.02, - 1.98)
In simple words: To find the interval form of a neighbourhood, you subtract and add the 'neighbourhood distance' (here, 0.02) from the center point (here, -2). So, it's \((-2 - 0.02, -2 + 0.02)\), which becomes \((-2.02, -1.98)\).

🎯 Exam Tip: Remember that a neighbourhood of 'a' with radius 'd' is represented as \((a - d, a + d)\). Always be careful with negative numbers.

 

Question 3. What is the interval form of \(|x - 5| < 0.25\)?
(a) (4.75, 5.25)
(b) (- 4.75, + 5.25)
(c) (- 5.25,-4.75)
(d) (- 5.25, 4.75)
Answer: (a) (4.75, 5.25)
In simple words: The expression \(|x - 5| < 0.25\) means that the distance between x and 5 is less than 0.25. This forms an open interval around 5, from \(5 - 0.25\) to \(5 + 0.25\). So, the interval is \((4.75, 5.25)\).

🎯 Exam Tip: Convert the modulus inequality \(|x - a| < d\) into its equivalent interval form \((a - d, a + d)\) by correctly identifying 'a' and 'd'.

 

Question 4. What is the interval form of \(|2x + 1| < \frac{1}{5}\)?
(a) \(\left(-\frac{6}{5},-\frac{4}{5}\right)\)
(b) \(\left(-\frac{6}{10},-\frac{4}{10}\right)\)
(c) \(\left(\frac{4}{10}, \frac{6}{10}\right)\)
(d) \(\left(-\frac{6}{10}, \frac{4}{10}\right)\)
Answer: (b) \(\left(-\frac{6}{10},-\frac{4}{10}\right)\)
In simple words: To change \(|2x + 1| < \frac{1}{5}\) to interval form, first divide by 2 to get \(|x + \frac{1}{2}| < \frac{1}{10}\). This means the center is \(-\frac{1}{2}\) and the radius is \(\frac{1}{10}\). So the interval is \((-\frac{1}{2} - \frac{1}{10}, -\frac{1}{2} + \frac{1}{10})\) which is \((-\frac{5}{10} - \frac{1}{10}, -\frac{5}{10} + \frac{1}{10})\) or \((-\frac{6}{10}, -\frac{4}{10})\).

🎯 Exam Tip: When the variable 'x' has a coefficient (e.g., \(2x\)), divide the entire inequality by that coefficient to get the standard form \(|x - a| < d\) before converting to an interval.

 

Question 5. What is the modulus form of N (5, 0.02) ?
(a) \(|x + 5| < 0.02\)
(b) \(|x - 0.02| < 5\)
(c) \(|x - 5| > 0.02\)
(d) \(|x - 5| < 0.02\)
Answer: (d) \(|x - 5| < 0.02\)
In simple words: The notation N(a, d) represents a neighbourhood with center 'a' and radius 'd'. Its modulus form is \(|x - a| < d\). Here, 'a' is 5 and 'd' is 0.02, so the modulus form is \(|x - 5| < 0.02\).

🎯 Exam Tip: Remember the direct conversion: N(center, radius) becomes \(|x - \text{center}| < \text{radius}\).

 

Question 6. If modulus form of N (a, 0.07) is \(|x - 10| < k\), then what will be the value of k?
(a) a
(b) 0.7
(c) 0.07
(d) 9.93
Answer: (c) 0.07
In simple words: A neighbourhood N(a, d) means the center is 'a' and the radius is 'd'. The modulus form \(|x - a| < d\) matches \(|x - 10| < k\). Comparing them, we see that 'a' is 10 and 'k' is 0.07.

🎯 Exam Tip: By comparing the given forms, you can directly identify the values of the parameters. In N(a, d), 'a' is the center and 'd' is the radius (k in this case).

 

Question 7. What is the value of \( \lim_{x \to 3} (3x - 1) \)?
(a) 9
(b) 10
(c) \(\frac{4}{3}\)
(d) 8
Answer: (d) 8
In simple words: To find the limit of \(3x - 1\) as x gets closer to 3, you just put 3 in place of x. So, \(3 \times 3 - 1 = 9 - 1 = 8\).

🎯 Exam Tip: For polynomial functions, direct substitution of the limit value into the function usually gives the answer. Only complex cases require advanced techniques.

 

Question 8. What is the value of \( \lim_{x \to 4} \sqrt{4x + 9} \)?
(a) 5
(b) 25
(c) \(\frac{7}{4}\)
(d) 7
Answer: (a) 5
In simple words: To find the limit of \(\sqrt{4x + 9}\) as x approaches 4, substitute x = 4 into the expression. This gives \(\sqrt{4 \times 4 + 9} = \sqrt{16 + 9} = \sqrt{25}\), which equals 5.

🎯 Exam Tip: Similar to polynomials, for continuous functions (like square roots of positive expressions), direct substitution is the primary method to evaluate limits.

 

Question 9. What is the value of \( \lim_{x \to 2} (10) \)?
(a) 10
(b)-2
(c) 8
(d) Indeterminate
Answer: (a) 10
In simple words: The limit of a constant number is always that same constant number, no matter what value x approaches. So, the limit of 10 is 10.

🎯 Exam Tip: The limit of a constant function \(f(x) = c\) as \(x \to a\) is always \(c\).

 

Question 10. What is the value of \( \lim_{x \to 3} \frac{x^4 - 81}{x - 3} \)?
(a) 192
(b) 324
(c) 36
(d) 108
Answer: (d) 108
In simple words: This limit is of the form \( \lim_{x \to a} \frac{x^n - a^n}{x - a} = n \cdot a^{n-1} \). Here, \(a = 3\) and \(n = 4\). So, the answer is \(4 \cdot 3^{4-1} = 4 \cdot 3^3 = 4 \cdot 27 = 108\).

🎯 Exam Tip: Recognize and apply standard limit formulas like \( \lim_{x \to a} \frac{x^n - a^n}{x - a} = n \cdot a^{n-1} \) to quickly solve such problems without extensive factorization.

 

Question 11. What is the value of \( \lim_{x \to -1} \frac{x^5 + 1}{x + 1} \)?
(a) - 5
(b) 5
(c) 4
(d) - 4
Answer: (b) 5
In simple words: This limit is of the form \( \lim_{x \to a} \frac{x^n - a^n}{x - a} = n \cdot a^{n-1} \). We can rewrite \(\frac{x^5 + 1}{x + 1}\) as \(\frac{x^5 - (-1)^5}{x - (-1)}\). Here, \(a = -1\) and \(n = 5\). So, the answer is \(5 \cdot (-1)^{5-1} = 5 \cdot (-1)^4 = 5 \cdot 1 = 5\).

🎯 Exam Tip: Be careful with negative values for 'a'. Remember that negative numbers raised to an even power are positive, and to an odd power are negative.

 

Question 12. If \(y = 10 - 3x\) and \(x \to -3\), then y tends to which value ?
(a) 1
(b) 9
(c) 19
(d) 7
Answer: (c) 19
In simple words: If \(y = 10 - 3x\) and x is approaching -3, just substitute -3 for x in the equation. So, \(y = 10 - 3 \times (-3) = 10 - (-9) = 10 + 9 = 19\).

🎯 Exam Tip: For linear functions, finding the value y tends to as x approaches a certain number is a direct substitution problem, as linear functions are continuous everywhere.

 

Section B

Answer the following questions in one sentence:

 

Question 1. Express 0.09 neighbourhood of 0 in interval form.
Answer: The 0.09 neighbourhood of 0 in interval form is \((0 - 0.09, 0 + 0.09)\), which simplifies to \((-0.09, 0.09)\).
In simple words: The neighbourhood of 0 with a radius of 0.09 means numbers between \(0 - 0.09\) and \(0 + 0.09\), giving the interval \((-0.09, 0.09)\).

🎯 Exam Tip: For a neighbourhood of 0, the interval form is always symmetric around 0, i.e., \((-d, d)\), where 'd' is the neighbourhood distance.

 

Question 2. Express 0.001 neighbourhood of - 5 in modulus form.
Answer: The 0.001 neighbourhood of -5 in modulus form is expressed by \(|x - (-5)| < 0.001\), which simplifies to \(|x + 5| < 0.001\).
In simple words: The modulus form \(|x - a| < d\) shows numbers within 'd' distance from 'a'. For a neighbourhood of -5 with 0.001 radius, 'a' is -5 and 'd' is 0.001, so it's \(|x - (-5)| < 0.001\).

🎯 Exam Tip: Pay close attention to negative signs. \(|x - (-a)|\) becomes \(|x + a|\).

 

Question 3. Express \(|x - 10| < \frac{1}{10}\) in neighbourhood form.
Answer: The modulus form \(|x - 10| < \frac{1}{10}\) in neighbourhood form is expressed by N \((10, \frac{1}{10})\).
In simple words: The expression \(|x - 10| < \frac{1}{10}\) means numbers are within \(\frac{1}{10}\) distance from 10. So, it's a neighbourhood (N) centered at 10 with a radius of \(\frac{1}{10}\).

🎯 Exam Tip: Remember that \(|x - a| < d\) directly translates to \(N(a, d)\) where 'a' is the center and 'd' is the radius.

 

Question 4. Express \(|2x| < \frac{1}{2}\) in interval form.
Answer: The inequality \(|2x| < \frac{1}{2}\) in interval form is expressed by \(\left(-\frac{1}{4}, \frac{1}{4}\right)\).
In simple words: To get x by itself, divide \(|2x| < \frac{1}{2}\) by 2, which gives \(|x| < \frac{1}{4}\). This means x is between \(-\frac{1}{4}\) and \(\frac{1}{4}\), so the interval is \((-\frac{1}{4}, \frac{1}{4})\).

🎯 Exam Tip: Always isolate \(|x|\) before determining the interval. If \(|x| < d\), the interval is \((-d, d)\).

 

Question 5. Express N (50, 0.8) in modulus form.
Answer: N (50, 0.8) in modulus form is expressed by \(|x - 50| < 0.8\).
In simple words: The neighbourhood N(50, 0.8) has a center of 50 and a radius of 0.8. In modulus form, this is written as \(|x - \text{center}| < \text{radius}\), so it becomes \(|x - 50| < 0.8\).

🎯 Exam Tip: Direct conversion from \(N(a, d)\) to \(|x - a| < d\) is a fundamental concept for this chapter.

 

Question 6. If N(a, 0.2) = \(|x - 7| < b\), then find the value of a.
Answer: Given N(a, 0.2) = \(|x - 7| < b\). By comparing the neighbourhood form N(a, d) with \(|x - a| < d\), we can see that 'a' in N(a, 0.2) corresponds to 7 in \(|x - 7|\). Therefore, the value of a is 7.
In simple words: The neighbourhood N(a, 0.2) has 'a' as its center. The modulus form \(|x - 7| < b\) shows its center is 7. So, 'a' must be 7.

🎯 Exam Tip: Match the components of the neighbourhood notation N(a, d) with the modulus form \(|x - a| < d\). The first value in N(a, d) is 'a', which is the center.

 

Question 7. If \(|x + 4| < 0.04 = (k, - 3.96)\), then find the value of k.
Answer: Given \(|x + 4| < 0.04\). This is \(|x - (-4)| < 0.04\). So, the center \(a = -4\) and the radius \(\delta = 0.04\).
The interval form is \((a - \delta, a + \delta)\).
Substituting the values, we get \((-4 - 0.04, -4 + 0.04)\) which is \((-4.04, -3.96)\).
Comparing this with \((k, -3.96)\), we find \(k = -4.04\).
In simple words: The expression \(|x + 4| < 0.04\) means x is between \(-4 - 0.04\) and \(-4 + 0.04\), which is \((-4.04, -3.96)\). Since the given interval is \((k, -3.96)\), then k must be \(-4.04\).

🎯 Exam Tip: Convert the modulus form to an interval form \((a - \delta, a + \delta)\) and then compare it with the given interval \((k, -3.96)\) to find 'k'. Remember to handle the center 'a' carefully if it's negative.

 

Question 8. Find the value of \( \lim_{x \to 5} (3x + 5) \).
Answer: \( \lim_{x \to 5} (3x + 5) = 3(5) + 5 = 15 + 5 = 20 \)
In simple words: To find the limit, simply replace x with 5 in the expression \(3x + 5\). This gives \(3 \times 5 + 5 = 15 + 5 = 20\).

🎯 Exam Tip: For simple polynomial functions, evaluating the limit involves direct substitution of the value that x approaches.

 

Question 9. Find the value of \( \lim_{x \to -3} \sqrt{2 - 2x} \).
Answer: \( \lim_{x \to -3} \sqrt{2 - 2x} = \sqrt{2 - 2(-3)} = \sqrt{2 + 6} = \sqrt{8} = 2\sqrt{2} \)
In simple words: Substitute -3 for x in the expression \(\sqrt{2 - 2x}\). This calculation leads to \(\sqrt{2 - (-6)} = \sqrt{2 + 6} = \sqrt{8}\), which simplifies to \(2\sqrt{2}\).

🎯 Exam Tip: Always perform direct substitution first. If the expression under the square root becomes negative, the limit might not exist in real numbers. Here, it remains positive.

 

Question 10. Find the value of \( \lim_{x \to 0} \frac{3x^2 - 4x + 10}{2x + 5} \)?
Answer: \( \lim_{x \to 0} \frac{3x^2 - 4x + 10}{2x + 5} = \frac{3(0)^2 - 4(0) + 10}{2(0) + 5} = \frac{0 - 0 + 10}{0 + 5} = \frac{10}{5} = 2 \)
In simple words: To find the limit, replace x with 0 in the fraction. This gives \(\frac{3(0)^2 - 4(0) + 10}{2(0) + 5}\), which simplifies to \(\frac{10}{5}\), which is 2.

🎯 Exam Tip: For rational functions, direct substitution works if the denominator does not become zero. If it becomes zero, further simplification (like factorization) is needed.

 

Question 11. Find the value of \( \lim_{x \to 2} \frac{x^5 - 32}{x - 2} \).
Answer: \( \lim_{x \to 2} \frac{x^5 - 32}{x - 2} \)
This is in the form \( \lim_{x \to a} \frac{x^n - a^n}{x - a} = n \cdot a^{n-1} \).
Here, \(a = 2\) and \(n = 5\).
So, the value is \(5 \cdot (2)^{5-1} = 5 \cdot 2^4 = 5 \cdot 16 = 80\).
In simple words: This limit matches a special formula: \( \lim_{x \to a} \frac{x^n - a^n}{x - a} = n \cdot a^{n-1} \). By using a=2 and n=5, we calculate \(5 \times 2^{(5-1)}\), which is \(5 \times 16\), giving 80.

🎯 Exam Tip: Recognize common limit forms to apply standard formulas, especially for indeterminate forms of type \(\frac{0}{0}\).

 

Question 12. Find the value of \( \lim_{x \to -a} \frac{x^m + a^m}{x + a} \), where m is an odd number.
Answer: Given \( \lim_{x \to -a} \frac{x^m + a^m}{x + a} \), where m is an odd number.
We can rewrite the expression as \( \lim_{x \to -a} \frac{x^m - (-a)^m}{x - (-a)} \). (Since m is odd, \(a^m = -(-a)^m\)).
This is in the form \( \lim_{x \to k} \frac{x^n - k^n}{x - k} = n \cdot k^{n-1} \).
Here, \(k = -a\) and \(n = m\).
So, the value is \(m(-a)^{m-1}\).
Since m is odd, \(m-1\) is an even number. Therefore, \((-a)^{m-1} = a^{m-1}\).
Thus, the limit is \(m \cdot a^{m-1}\).
In simple words: When m is an odd number, \(a^m\) is the same as \(-(-a)^m\). So, we can change the expression to fit the limit formula \( \lim_{x \to k} \frac{x^n - k^n}{x - k} = n \cdot k^{n-1} \). With \(k = -a\) and \(n = m\), the answer becomes \(m(-a)^{m-1}\). Since \(m-1\) is even, \((-a)^{m-1}\) becomes \(a^{m-1}\), making the final answer \(m a^{m-1}\).

🎯 Exam Tip: Be mindful of algebraic manipulations, especially with negative bases and odd/even exponents, to fit expressions into standard limit formulas.

 

Question 13. If \( \lim_{x \to -1} (4x + k) = 6 \), then find the value of k.
Answer: Given \( \lim_{x \to -1} (4x + k) = 6 \).
By substituting \(x = -1\) into the expression, we get:
\(4(-1) + k = 6\)
\(-4 + k = 6\)
\(k = 6 + 4\)
\(k = 10\)
In simple words: We are given that the limit of \(4x + k\) as x approaches -1 is 6. By putting -1 in place of x, we get \(-4 + k = 6\). Solving this simple equation for k gives \(k = 10\).

🎯 Exam Tip: If the limit of a polynomial is known, direct substitution is used to find any unknown constants. This type of problem often appears in basic limit evaluations.

 

Question 14. If \( \lim_{x \to 3} \frac{2}{3x + k} = \frac{1}{7} \), then find the value of k.
Answer: Given \( \lim_{x \to 3} \frac{2}{3x + k} = \frac{1}{7} \).
By substituting \(x = 3\) into the expression, we get:
\( \frac{2}{3(3) + k} = \frac{1}{7} \)
\( \frac{2}{9 + k} = \frac{1}{7} \)
Cross-multiplying:
\(2 \times 7 = 1 \times (9 + k)\)
\(14 = 9 + k\)
\(k = 14 - 9\)
\(k = 5\)
In simple words: We know the limit of \(\frac{2}{3x + k}\) as x goes to 3 is \(\frac{1}{7}\). Put \(x = 3\) into the fraction: \(\frac{2}{9 + k} = \frac{1}{7}\). Then, cross-multiply to solve for k: \(14 = 9 + k\), which means \(k = 5\).

🎯 Exam Tip: For rational functions, if the limit exists and the denominator is not zero after substitution, you can directly substitute the limit value to form an equation and solve for the unknown constant.

 

Section C

Answer the following questions as required:

 

Question 1. Define an open interval.
Answer: If 'a' and 'b' are real numbers with \(a < b\), then the set of all real numbers between 'a' and 'b', not including 'a' and 'b' themselves, is called an open interval. It is denoted by \((a, b)\). Thus, \((a, b) = \{x | a < x < b, x \in R\}\).
In simple words: An open interval includes all numbers between two points, but not the points themselves. For example, \((2, 5)\) means all numbers greater than 2 but less than 5.

🎯 Exam Tip: Clearly state the conditions for 'a' and 'b' (\(a, b \in R\) and \(a < b\)), and accurately represent the notation and set definition.

 

Question 2. Define the \(\delta\) neighbourhood of a.
Answer: If 'a' is a real number and \(\delta\) (delta) is a non-negative real number, then the open interval \((a - \delta, a + \delta)\) is called the \(\delta\) neighbourhood of 'a'. It is denoted by N \((a, \delta)\). Therefore, N \((a, \delta) = \{x | (a - \delta) < x < (a + \delta), x \in R\}\) or equivalently, \( \{x | |x - a| < \delta, x \in R\} \).
In simple words: A \(\delta\) neighbourhood of 'a' means all numbers 'x' that are very close to 'a', within a distance of \(\delta\). It's an open interval \((a - \delta, a + \delta)\), or numbers 'x' where the distance \(|x - a|\) is less than \(\delta\).

🎯 Exam Tip: Be precise with the definition, including the conditions for 'a' and \(\delta\), and present both the interval notation and the modulus inequality form.

 

Question 3. Define the punctured \(\delta\) neighbourhood of a.
Answer: If 'a' is a real number and \(\delta\) is a non-negative real number, then the open interval \((a - \delta, a + \delta)\) with the point 'a' removed, i.e., \((a - \delta, a + \delta) - \{a\}\), is called the punctured \(\delta\) neighbourhood of 'a'. It is denoted by N* \((a, \delta)\). Thus, N* \((a, \delta) = N(a, \delta) - \{a\}\) which means \( \{x | (a - \delta) < x < (a + \delta); x \ne a, x \in R\} \) or equivalently, \( \{x | |x - a| < \delta; x \ne a, x \in R\} \).
In simple words: A punctured \(\delta\) neighbourhood is like a regular neighbourhood \((a - \delta, a + \delta)\), but it specifically excludes the center point 'a' itself. So, 'x' can be any number close to 'a' but not 'a'.

🎯 Exam Tip: The key difference between a neighbourhood and a punctured neighbourhood is the exclusion of the center point 'a'. Make sure to include \(x \ne a\) in your definition.

 

Question 4. Express the interval form \((-0.5, 0.5)\) in modulus form.
Answer: Given interval form is \((-0.5, 0.5)\).
For an interval \((a - \delta, a + \delta)\), we have:
\(a - \delta = -0.5\)
\(a + \delta = 0.5\)
Adding the two equations: \((a - \delta) + (a + \delta) = -0.5 + 0.5 \implies 2a = 0 \implies a = 0\).
Substitute \(a = 0\) into \(a + \delta = 0.5 \implies 0 + \delta = 0.5 \implies \delta = 0.5\).
The modulus form is \(|x - a| < \delta\).
Substituting \(a = 0\) and \(\delta = 0.5\), we get \(|x - 0| < 0.5\), which simplifies to \(|x| < 0.5\).
In simple words: We have the interval \((-0.5, 0.5)\). The middle point (center) is 0, and the distance from the center to either end (radius) is 0.5. So, the modulus form is \(|x - 0| < 0.5\), which is simply \(|x| < 0.5\).

🎯 Exam Tip: To find 'a' and \(\delta\) from an interval \((a - \delta, a + \delta)\), use the formulas \(a = \frac{(a-\delta) + (a+\delta)}{2}\) and \(\delta = \frac{(a+\delta) - (a-\delta)}{2}\).

 

Question 5. Express the interval form \((-8.75, -7.25)\) in neighbourhood form.
Answer: Given interval form is \((-8.75, -7.25)\).
For an interval \((a - \delta, a + \delta)\), we have:
\(a - \delta = -8.75\)
\(a + \delta = -7.25\)
Adding the two equations: \(2a = -8.75 + (-7.25) \implies 2a = -16 \implies a = -8\).
Substitute \(a = -8\) into \(a + \delta = -7.25 \implies -8 + \delta = -7.25 \implies \delta = -7.25 + 8 \implies \delta = 0.75\).
The neighbourhood form is N \((a, \delta)\).
Substituting \(a = -8\) and \(\delta = 0.75\), we get N \((-8, 0.75)\).
In simple words: For the interval \((-8.75, -7.25)\), the center 'a' is found by adding the two ends and dividing by 2: \(\frac{-8.75 + (-7.25)}{2} = \frac{-16}{2} = -8\). The radius \(\delta\) is found by subtracting the center from the upper end: \(-7.25 - (-8) = 0.75\). So, the neighbourhood is N \((-8, 0.75)\).

🎯 Exam Tip: Similar to Question 4, use the midpoint formula for 'a' and the distance from the midpoint to an endpoint for \(\delta\). Be careful with arithmetic involving negative numbers.

 

Question 6. If N \((k_1, 0.5) = (19.5, k_2)\), then find the values of \(k_1\) and \(k_2\).
Answer: Given N \((k_1, 0.5) = (19.5, k_2)\).
From the definition of a neighbourhood N \((a, \delta)\), we know that 'a' is the center and \(\delta\) is the radius. The interval form is \((a - \delta, a + \delta)\).
Comparing N \((k_1, 0.5)\) with N \((a, \delta)\), we have \(a = k_1\) and \(\delta = 0.5\).
The interval form is \((k_1 - 0.5, k_1 + 0.5)\).
Comparing this with the given interval \((19.5, k_2)\), we have:
\(k_1 - 0.5 = 19.5 \implies k_1 = 19.5 + 0.5 \implies k_1 = 20\).
And \(k_1 + 0.5 = k_2\).
Substitute \(k_1 = 20\) into the second equation:
\(20 + 0.5 = k_2 \implies k_2 = 20.5\).
Thus, \(k_1 = 20\) and \(k_2 = 20.5\).
In simple words: In N \((k_1, 0.5)\), \(k_1\) is the center and 0.5 is the radius. The interval \((19.5, k_2)\) comes from \((k_1 - 0.5, k_1 + 0.5)\). So, \(k_1 - 0.5 = 19.5\), which gives \(k_1 = 20\). Then, \(k_2 = k_1 + 0.5 = 20 + 0.5 = 20.5\).

🎯 Exam Tip: Break down the neighbourhood notation into its components (center and radius), form equations from the given interval, and solve for the unknown variables. This requires a strong grasp of the definition of a neighbourhood.

 

Question 7. Express \(|3x + 1| < 2\) in neighbourhood and interval form.
Answer: Given modulus form is \(|3x + 1| < 2\).
To convert to standard modulus form \(|x - a| < \delta\), divide the inequality by 3:
\( \left|\frac{3x + 1}{3}\right| < \frac{2}{3} \)
\( \left|x + \frac{1}{3}\right| < \frac{2}{3} \)
This is \( \left|x - \left(-\frac{1}{3}\right)\right| < \frac{2}{3} \).
So, \(a = -\frac{1}{3}\) and \(\delta = \frac{2}{3}\).

In neighbourhood form: N \((a, \delta)\) is N \(\left(-\frac{1}{3}, \frac{2}{3}\right)\).

In interval form: \((a - \delta, a + \delta)\) is \(\left(-\frac{1}{3} - \frac{2}{3}, -\frac{1}{3} + \frac{2}{3}\right)\).
This simplifies to \(\left(-\frac{3}{3}, \frac{1}{3}\right)\) which is \(\left(-1, \frac{1}{3}\right)\).
In simple words: First, rewrite \(|3x + 1| < 2\) as \(|x + \frac{1}{3}| < \frac{2}{3}\). This means the center 'a' is \(-\frac{1}{3}\) and the radius \(\delta\) is \(\frac{2}{3}\). So, the neighbourhood form is N \((-\frac{1}{3}, \frac{2}{3})\). For the interval form, it's \((a - \delta, a + \delta)\), which is \((-\frac{1}{3} - \frac{2}{3}, -\frac{1}{3} + \frac{2}{3})\), simplifying to \((-1, \frac{1}{3})\).

🎯 Exam Tip: Always transform the given modulus inequality into the standard form \(|x - a| < \delta\) first, by dividing by the coefficient of x, to correctly identify the center 'a' and radius \(\delta\).

 

Question 8. If \(|x - A_1| < 0.09 = (A_2, 4.09)\), then find the values of \(A_1\) and \(A_2\).
Answer: Given \(|x - A_1| < 0.09\).
This modulus form represents a neighbourhood with center \(A_1\) and radius \(\delta = 0.09\).
The corresponding interval form is \((A_1 - 0.09, A_1 + 0.09)\).
We are also given that the interval is \((A_2, 4.09)\).
Comparing the two interval forms:
\(A_1 - 0.09 = A_2\) (Equation 1)
\(A_1 + 0.09 = 4.09\) (Equation 2)
From Equation 2:
\(A_1 = 4.09 - 0.09 \implies A_1 = 4\).
Substitute \(A_1 = 4\) into Equation 1:
\(4 - 0.09 = A_2 \implies A_2 = 3.91\).
Therefore, \(A_1 = 4\) and \(A_2 = 3.91\).
In simple words: The modulus \(|x - A_1| < 0.09\) means the center is \(A_1\) and the radius is 0.09. This makes the interval \((A_1 - 0.09, A_1 + 0.09)\). We are told this interval is \((A_2, 4.09)\). So, \(A_1 + 0.09 = 4.09\), which means \(A_1 = 4\). Then, \(A_2 = A_1 - 0.09 = 4 - 0.09 = 3.91\).

🎯 Exam Tip: This question combines understanding of modulus form, interval form, and solving simultaneous equations. Identify the center and radius from the modulus form, set up equations based on the given interval, and solve them carefully.

 

Question 9. Explain the meaning of \(x \to a\).
Answer: The expression \(x \to a\) means that the value of the variable 'x' is approaching the number 'a' very closely. This can happen by increasing 'x' towards 'a' from a smaller value (left-hand side) or by decreasing 'x' towards 'a' from a larger value (right-hand side). It is crucial to note that while x gets very close to 'a', \(x \ne a\).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दिखाता है कि कैसे 'x' मान 'a' के करीब आता है. बाईं ओर से (छोटे मानों से) 'x' बढ़ते हुए 'a' तक पहुँचता है (0.9, 0.99, 0.999), और दाईं ओर से (बड़े मानों से) 'x' घटते हुए 'a' तक पहुँचता है (1.1, 1.01, 1.001). दोनों मामलों में, 'x' 'a' के बहुत नज़दीक आता है, लेकिन 'a' के बराबर नहीं होता है.

In simple words: \(x \to a\) means x is getting extremely close to the number 'a', from both sides (smaller or larger values), but it never actually becomes equal to 'a'.

🎯 Exam Tip: Emphasize that \(x\) approaches \(a\) from both sides but does not necessarily equal \(a\). This distinction is fundamental to the concept of limits.

 

Question 10. Explain the meaning of \(x \to 0\).
Answer: The notation \(x \to 0\) signifies that the value of the variable 'x' is approaching zero very closely. This can occur either by decreasing positive values of 'x' (approaching from the right-hand side) or by increasing negative values of 'x' (approaching from the left-hand side). The key point is that 'x' gets arbitrarily close to 0 but never actually equals 0.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दर्शाता है कि 'x' मान कैसे 0 के करीब आता है. बाईं ओर से (नकारात्मक मानों से) 'x' बढ़ते हुए 0 तक पहुँचता है (-0.1, -0.01, -0.001), और दाईं ओर से (सकारात्मक मानों से) 'x' घटते हुए 0 तक पहुँचता है (0.1, 0.01, 0.001). दोनों मामलों में, 'x' 0 के बहुत नज़दीक आता है, लेकिन 0 के बराबर नहीं होता है.

In simple words: \(x \to 0\) means that the value of x is getting extremely close to zero, either from positive numbers or negative numbers, but it is not exactly zero.

🎯 Exam Tip: Similar to \(x \to a\), for \(x \to 0\), highlight that \(x\) approaches 0 from both positive and negative sides but is never equal to 0. This is a special case of the general limit definition.

 

Question 11. Define limit of a function.
Answer: The limit of a function \(f(x)\) describes the value that \(f(x)\) approaches as the input 'x' gets closer and closer to some number 'a'. If, as 'x' approaches 'a' (but not necessarily equals 'a'), the value of \(f(x)\) gets closer and closer to a definite number 'l', then 'l' is called the limit of the function \(f(x)\) as 'x' tends to 'a'. This is written symbolically as \( \lim_{x \to a} f(x) = l \). It means that \(f(x)\) is approaching 'l', but \(f(x)\) does not have to be equal to 'l' at \(x=a\).
In simple words: The limit of a function tells us what value the function output gets very, very close to, as the input gets very, very close to a specific number. It's like predicting the function's height at a point, even if there's a hole there.

🎯 Exam Tip: The definition must clearly state that \(x\) approaches \(a\) but \(x \neq a\), and \(f(x)\) approaches \(l\) but \(f(x) \neq l\). The concept of 'approaching' is central.

 

Question 12. State multiplication working rule of limit.
Answer: If \(f(x)\) and \(g(x)\) are two functions of a real variable 'x', and if their limits exist as \(x \to a\), such that \( \lim_{x \to a} f(x) = l \) and \( \lim_{x \to a} g(x) = m \), then the multiplication working rule for limits states that the limit of the product of these two functions is equal to the product of their individual limits. This is expressed as:
\[ \lim_{x \to a} [f(x) \cdot g(x)] = \left( \lim_{x \to a} f(x) \right) \cdot \left( \lim_{x \to a} g(x) \right) = l \cdot m \]
In simple words: If you want to find the limit of two functions multiplied together, you can find the limit of each function separately and then multiply those two limits.

🎯 Exam Tip: State the rule formally with conditions that the individual limits must exist. The formula \( \lim [f(x) \cdot g(x)] = \lim f(x) \cdot \lim g(x) \) is essential.

 

Question 13. State division working rule of limit.
Answer: If \(f(x)\) and \(g(x)\) are two functions of a real variable 'x', and if their limits exist as \(x \to a\), such that \( \lim_{x \to a} f(x) = l \) and \( \lim_{x \to a} g(x) = m \), where \(m \ne 0\), then the division working rule for limits states that the limit of the division of these two functions is equal to the division of their individual limits. This is expressed as:
\[ \lim_{x \to a} \left[ \frac{f(x)}{g(x)} \right] = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)} = \frac{l}{m} \]
The critical condition for this rule is that the limit of the function in the denominator, \(m\), must not be zero.
In simple words: If you need to find the limit of one function divided by another, you can find the limit of the top function and divide it by the limit of the bottom function. But, the limit of the bottom function cannot be zero.

🎯 Exam Tip: The non-zero denominator condition (\(m \neq 0\)) is extremely important for the division rule. Always include it in the definition.

 

Question 14. State the standard form of limit of a polynomial.
Answer: Suppose \(f(x)\) is a polynomial function given by \(f(x) = a_0 + a_1x + a_2x^2 + \dots + a_nx^n\). The standard form of the limit of this polynomial as \(x \to b\) is found by direct substitution of 'b' for 'x' into the polynomial. This is because polynomial functions are continuous everywhere. So, the standard form is:
\[ \lim_{x \to b} f(x) = \lim_{x \to b} (a_0 + a_1x + a_2x^2 + \dots + a_nx^n) = a_0 + a_1b + a_2b^2 + \dots + a_nb^n \]
In simple words: To find the limit of a polynomial function as x gets close to a number, you just replace x with that number in the polynomial. Polynomials are "smooth" functions, so this direct substitution always works.

🎯 Exam Tip: For any polynomial function, the limit as \(x \to b\) is simply \(f(b)\) due to the property of continuity. This is a straightforward rule.

 

Section D

Find the values of the following:

 

Question 1. \( \lim_{x \to 1} \frac{3x^2 - 4x + 1}{x - 1} \).
Answer: Given limit: \( \lim_{x \to 1} \frac{3x^2 - 4x + 1}{x - 1} \).
If we substitute \(x = 1\), we get \(\frac{3(1)^2 - 4(1) + 1}{1 - 1} = \frac{3 - 4 + 1}{0} = \frac{0}{0}\), which is an indeterminate form.
We need to factorize the numerator to find the common factor \((x - 1)\).
\( \lim_{x \to 1} \frac{3x^2 - 3x - x + 1}{x - 1} \)
\( = \lim_{x \to 1} \frac{3x(x - 1) - 1(x - 1)}{x - 1} \)
\( = \lim_{x \to 1} \frac{(3x - 1)(x - 1)}{x - 1} \)
Since \(x \to 1\), \(x \ne 1\), so we can cancel out \((x - 1)\):
\( = \lim_{x \to 1} (3x - 1) \)
Now, substitute \(x = 1\):
\( = 3(1) - 1 = 3 - 1 = 2 \).
Hence, \( \lim_{x \to 1} \frac{3x^2 - 4x + 1}{x - 1} = 2 \).
In simple words: If we put \(x = 1\), we get \(\frac{0}{0}\), which means we need to simplify. We factor the top part \((3x^2 - 4x + 1)\) into \((3x - 1)(x - 1)\). Then, we can cancel \((x - 1)\) from the top and bottom. What's left is \(3x - 1\). Now, putting \(x = 1\) into \(3x - 1\) gives \(3(1) - 1 = 2\).

🎯 Exam Tip: When direct substitution results in an indeterminate form like \(\frac{0}{0}\), factorize the numerator and/or denominator to cancel common factors before re-evaluating the limit.

 

Question 2. \( \lim_{x \to 3} \frac{x - 3}{2x^2 - 3x - 9} \).
Answer: Given limit: \( \lim_{x \to 3} \frac{x - 3}{2x^2 - 3x - 9} \).
If we substitute \(x = 3\), we get \(\frac{3 - 3}{2(3)^2 - 3(3) - 9} = \frac{0}{18 - 9 - 9} = \frac{0}{0}\), which is an indeterminate form.
We need to factorize the denominator to find the common factor \((x - 3)\).
\( \lim_{x \to 3} \frac{x - 3}{2x^2 - 6x + 3x - 9} \)
\( = \lim_{x \to 3} \frac{x - 3}{2x(x - 3) + 3(x - 3)} \)
\( = \lim_{x \to 3} \frac{x - 3}{(2x + 3)(x - 3)} \)
Since \(x \to 3\), \(x \ne 3\), so we can cancel out \((x - 3)\):
\( = \lim_{x \to 3} \frac{1}{2x + 3} \)
Now, substitute \(x = 3\):
\( = \frac{1}{2(3) + 3} = \frac{1}{6 + 3} = \frac{1}{9} \).
Hence, \( \lim_{x \to 3} \frac{x - 3}{2x^2 - 3x - 9} = \frac{1}{9} \).
In simple words: If we put \(x = 3\), we get \(\frac{0}{0}\), so we need to simplify. Factor the bottom part \((2x^2 - 3x - 9)\) into \((2x + 3)(x - 3)\). Then, cancel \((x - 3)\) from the top and bottom. What's left is \(\frac{1}{2x + 3}\). Now, putting \(x = 3\) into this gives \(\frac{1}{2(3) + 3} = \frac{1}{9}\).

🎯 Exam Tip: Always look for common factors to cancel in indeterminate forms. Factoring quadratic expressions is a key skill here.

 

Question 3. \( \lim_{x \to -1} \frac{3x^2 - 2x - 5}{x + 1} \).
Answer: Given limit: \( \lim_{x \to -1} \frac{3x^2 - 2x - 5}{x + 1} \).
If we substitute \(x = -1\), we get \(\frac{3(-1)^2 - 2(-1) - 5}{-1 + 1} = \frac{3 + 2 - 5}{0} = \frac{0}{0}\), which is an indeterminate form.
We need to factorize the numerator to find the common factor \((x + 1)\).
\( \lim_{x \to -1} \frac{3x^2 - 5x + 3x - 5}{x + 1} \)
\( = \lim_{x \to -1} \frac{x(3x - 5) + 1(3x - 5)}{x + 1} \)
\( = \lim_{x \to -1} \frac{(x + 1)(3x - 5)}{x + 1} \)
Since \(x \to -1\), \(x \ne -1\), so we can cancel out \((x + 1)\):
\( = \lim_{x \to -1} (3x - 5) \)
Now, substitute \(x = -1\):
\( = 3(-1) - 5 = -3 - 5 = -8 \).
Hence, \( \lim_{x \to -1} \frac{3x^2 - 2x - 5}{x + 1} = -8 \).
In simple words: If we put \(x = -1\), we get \(\frac{0}{0}\). So, we factor the top part \((3x^2 - 2x - 5)\) into \((x + 1)(3x - 5)\). Then, we cancel \((x + 1)\) from top and bottom. This leaves \(3x - 5\). Now, putting \(x = -1\) into \(3x - 5\) gives \(3(-1) - 5 = -3 - 5 = -8\).

🎯 Exam Tip: For indeterminate forms, factorization is key. Remember that if \(x \to -a\), then \((x+a)\) is often a common factor. Look for ways to split the middle term in the numerator.

 

Question 4. \( \lim_{x \to 1} \frac{x^2 + 2x - 3}{x^2 - 1} \).
Answer: Given limit: \( \lim_{x \to 1} \frac{x^2 + 2x - 3}{x^2 - 1} \).
If we substitute \(x = 1\), we get \(\frac{1^2 + 2(1) - 3}{1^2 - 1} = \frac{1 + 2 - 3}{1 - 1} = \frac{0}{0}\), which is an indeterminate form.
We need to factorize both the numerator and the denominator to find the common factor \((x - 1)\).
Numerator: \(x^2 + 2x - 3 = x^2 + 3x - x - 3 = x(x + 3) - 1(x + 3) = (x - 1)(x + 3)\).
Denominator: \(x^2 - 1 = (x - 1)(x + 1)\).
So the limit becomes:
\( = \lim_{x \to 1} \frac{(x - 1)(x + 3)}{(x - 1)(x + 1)} \)
Since \(x \to 1\), \(x \ne 1\), so we can cancel out \((x - 1)\):
\( = \lim_{x \to 1} \frac{x + 3}{x + 1} \)
Now, substitute \(x = 1\):
\( = \frac{1 + 3}{1 + 1} = \frac{4}{2} = 2 \).
Hence, \( \lim_{x \to 1} \frac{x^2 + 2x - 3}{x^2 - 1} = 2 \).
In simple words: When \(x = 1\), both top and bottom are 0, so we factor them. The top \((x^2 + 2x - 3)\) factors to \((x - 1)(x + 3)\). The bottom \((x^2 - 1)\) factors to \((x - 1)(x + 1)\). We cancel \((x - 1)\) from both. Then we are left with \(\frac{x + 3}{x + 1}\). Putting \(x = 1\) in this gives \(\frac{1 + 3}{1 + 1} = \frac{4}{2} = 2\).

🎯 Exam Tip: Always factor both the numerator and denominator completely to identify all common factors that lead to the \(\frac{0}{0}\) form. Remember \(a^2 - b^2 = (a-b)(a+b)\).

 

Question 5. Find the value of \( \lim_{x\rightarrow \frac{1}{2}} \frac{2x^2+5x-3}{4x^2-1} \).
Answer: The limit of the given function is found by first simplifying the expression. Factor the numerator into \((2x-1)(x+3)\) and the denominator into \((2x-1)(2x+1)\). The common factor \((2x-1)\) is then canceled out. After cancellation, substitute \(x = \frac{1}{2}\) into the simplified expression \(\frac{x+3}{2x+1}\) to get the final value of \(\frac{7}{4}\).
In simple words: First, simplify the complex fraction by factoring the top and bottom parts and canceling out the common term. Then, put the value \(x = \frac{1}{2}\) into the simplified math problem to get the answer.

🎯 Exam Tip: Always look for common factors to simplify rational functions before substituting the limit value, especially if direct substitution leads to an indeterminate form (like 0/0).

 

Question 6. Find the value of \( \lim_{x\rightarrow -3} \frac{2x^2+9x+9}{2x^2+7x+3} \).
Answer: To find the limit, simplify the fraction. The numerator can be factored as \((2x+3)(x+3)\) and the denominator as \((2x+1)(x+3)\). After canceling the common factor \((x+3)\), substitute \(x = -3\) into the remaining expression \(\frac{2x+3}{2x+1}\) to calculate the limit, which is \(\frac{3}{5}\).
In simple words: Factor the top and bottom parts of the fraction and remove any common parts. Then, put \(x = -3\) into what's left to find the answer.

🎯 Exam Tip: Factoring quadratic expressions correctly is crucial. Remember that if \(x=a\) is a root, then \((x-a)\) is a factor.

 

Question 7. Find the value of \( \lim_{x\rightarrow -\frac{1}{2}} \frac{2x^2+3x+1}{2x^2-x-1} \).
Answer: To evaluate the limit, first factor the numerator into \((2x+1)(x+1)\) and the denominator into \((2x+1)(x-1)\). Then, cancel the common factor \((2x+1)\). Finally, replace \(x\) with \(-\frac{1}{2}\) in the simplified expression \(\frac{x+1}{x-1}\) to find the limit, which is \(-\frac{1}{3}\).
In simple words: Break down the top and bottom of the fraction into simpler multiplication parts and cancel anything that is the same. After that, replace \(x\) with \(-\frac{1}{2}\) to get the final number.

🎯 Exam Tip: When dealing with fractional limits, pay extra attention to signs and fraction arithmetic during substitution.

 

Question 8. Find the value of \( \lim_{x\rightarrow 0} \frac{1}{x} \left[ \frac{5x+14}{3x+7} - 2 \right] \).
Answer: The limit is found by simplifying the expression inside the brackets first. Combine the terms \(\frac{5x+14}{3x+7}\) and \(-2\) into a single fraction. This simplifies to \(\frac{-x}{3x+7}\). Then, multiply by \(\frac{1}{x}\) and cancel the common factor \(x\). Finally, substitute \(x=0\) into the resulting expression \(\frac{-1}{3x+7}\) to get the limit, which is \(-\frac{1}{7}\).
In simple words: First, combine the numbers inside the square brackets into one fraction. Then, multiply this new fraction by \(\frac{1}{x}\) and remove the \(x\) from top and bottom. Finally, put \(x=0\) into the simple fraction to get the answer.

🎯 Exam Tip: Always simplify expressions fully before attempting to substitute the limit value, especially when a variable is in the denominator leading to a \(\frac{0}{0}\) form.

 

Question 9. Find the value of \( \lim_{x\rightarrow 0} \frac{1}{x} \left[ \frac{5x+14}{x+7} - 2 \right] \).
Answer: The limit is calculated by first simplifying the expression inside the brackets. Combine the fraction \(\frac{5x+14}{3x+7}\) with \(-2\) into a single fraction. This simplifies to \(\frac{-x}{3x+7}\). Next, multiply by \(\frac{1}{x}\) and cancel the common factor \(x\). Finally, substitute \(x=0\) into the resulting expression \(\frac{-1}{3x+7}\) to determine the limit, which is \(-\frac{1}{7}\).
In simple words: First, make the parts inside the brackets into one simple fraction. Then, multiply by \(\frac{1}{x}\) and remove the matching \(x\) from the top and bottom. Lastly, put \(x=0\) into the final simple fraction to get the answer.

🎯 Exam Tip: When fractions are combined, ensure the common denominator is found correctly. This prevents errors in simplification and final substitution.

 

Question 10. Find the value of \( \lim_{x\rightarrow 2} \left[ \frac{2}{x-2} - \frac{4}{x^2-2x} \right] \).
Answer: To find the limit, first simplify the expression inside the brackets by combining the two fractions. Factor the denominator \(x^2-2x\) as \(x(x-2)\) to find a common denominator. Then, combine the fractions into \(\frac{2x-4}{x(x-2)}\). Factor \(2x-4\) as \(2(x-2)\) and cancel the common factor \((x-2)\). Finally, substitute \(x=2\) into the simplified expression \(\frac{2}{x}\) to get the limit, which is \(1\).
In simple words: Combine the two fractions inside the brackets into one by finding a common bottom part. Simplify the top part and then remove any common terms from top and bottom. After that, put \(x=2\) into the simple fraction to get the answer.

🎯 Exam Tip: When combining algebraic fractions, always factor denominators to easily identify the least common multiple.

 

Question 11. Find the value of \( \lim_{x\rightarrow 0} 1 + \frac{2}{3+\frac{4}{x}} \).
Answer: To evaluate this limit, first simplify the complex fraction by multiplying the numerator and denominator of \(\frac{2}{3+\frac{4}{x}}\) by \(x\). This simplifies to \(\frac{2x}{3x+4}\). Then, substitute \(x=0\) into the simplified expression \(1 + \frac{2x}{3x+4}\) to obtain the final limit, which is \(1\).
In simple words: First, simplify the fraction within a fraction by multiplying the top and bottom by \(x\). Then, put \(x=0\) into the simplified sum to find the answer.

🎯 Exam Tip: Simplify complex fractions step-by-step to avoid errors. Multiplying by a common factor in the nested denominator helps simplify efficiently.

 

Question 12. Find the value of \( \lim_{x\rightarrow -p} \frac{x^4-p^4}{x^3+p^3} \).
Answer: To evaluate the limit, factor both the numerator and the denominator. The numerator \(x^4-p^4\) factors into \((x^2+p^2)(x+p)(x-p)\) and the denominator \(x^3+p^3\) factors into \((x+p)(x^2-xp+p^2)\). Cancel the common factor \((x+p)\). Then, substitute \(x=-p\) into the simplified expression \(\frac{(x^2+p^2)(x-p)}{x^2-xp+p^2}\) to find the limit, which is \(-\frac{4}{3}p\).
In simple words: Break down the top and bottom of the fraction into multiplication parts using formulas like difference of squares and sum of cubes. Cancel the matching parts. Then, put \(x=-p\) into the remaining part to get the final answer.

🎯 Exam Tip: Recognize and correctly apply algebraic factorization formulas (e.g., \(a^2-b^2\), \(a^3+b^3\)) for higher powers to simplify expressions.

 

Question 13. Find the value of \( \lim_{x\rightarrow -a} \frac{x^m+a^m}{x+a} \) where m is an odd number.
Answer: [Answer content missing in source document.]
In simple words: [Explanation content missing in source document.]

🎯 Exam Tip: [Exam Tip content missing in source document.]

 

Question 14. Find the value of \( \lim_{x\rightarrow -2} \frac{x^{10}-1024}{x^5+32} \).
Answer: To find the limit, factor the numerator \(x^{10}-1024\) as a difference of squares: \((x^5)^2 - (32)^2\), which becomes \((x^5-32)(x^5+32)\). Then, cancel the common factor \((x^5+32)\) from the numerator and denominator. Substitute \(x=-2\) into the remaining expression \((x^5-32)\) to get the final limit, which is \(-64\).
In simple words: Break down the top part of the fraction using the difference of squares rule. Then, remove the matching part from the top and bottom. After that, put \(x=-2\) into the remaining math problem to find the answer.

🎯 Exam Tip: When numbers are powers of 2, recognize patterns like \(1024 = 2^{10}\) and \(32 = 2^5\), which helps in factoring expressions.

 

Question 15. Find the value of \( \lim_{x\rightarrow -1} \frac{x^{2017}+1}{x^{2018}-1} \).
Answer: This limit can be found using the standard limit formula \(\lim_{x \rightarrow a} \frac{x^n-a^n}{x-a} = n \cdot a^{n-1}\). Rewrite the numerator as \(x^{2017}-(-1)^{2017}\) and the denominator as \(x^{2018}-(-1)^{2018}\). Then, divide both numerator and denominator by \((x-(-1))\) and apply the formula. This gives \(\frac{2017(-1)^{2016}}{2018(-1)^{2017}}\). Since \((-1)^{2016}=1\) and \((-1)^{2017}=-1\), the expression simplifies to \(\frac{2017}{2018} \times \frac{1}{-1}\), yielding the result \(-\frac{2017}{2018}\).
In simple words: Use a special math rule for limits of powers. Change the numbers in the fraction to fit this rule. Then, apply the rule to both the top and bottom, calculate the powers of -1, and simplify to get the answer.

🎯 Exam Tip: Remember that \((-1)\) raised to an even power is \(1\), and \((-1)\) raised to an odd power is \(-1\). This is critical for simplifying such expressions.

 

Question 16. Find the value of \( \lim_{x\rightarrow 1} \frac{x^{\frac{7}{2}}-1}{x^{\frac{3}{2}}-1} \).
Answer: This limit is solved using the standard limit formula \(\lim_{x \rightarrow a} \frac{x^n-a^n}{x-a} = n \cdot a^{n-1}\). Divide both the numerator and denominator by \((x-1)\). Then, apply the formula to both parts, noting that \(a=1\). This results in \(\frac{\frac{7}{2}(1)^{\frac{7}{2}-1}}{\frac{3}{2}(1)^{\frac{3}{2}-1}}\), which simplifies to \(\frac{7}{2} \div \frac{3}{2}\), giving the final answer \(\frac{7}{3}\).
In simple words: Use a specific limit rule for expressions with powers, treating the top and bottom of the fraction separately. Since \(1\) raised to any power is still \(1\), the problem simplifies quickly to just dividing the power numbers.

🎯 Exam Tip: The formula \(\lim_{x \rightarrow a} \frac{x^n-a^n}{x-a} = n \cdot a^{n-1}\) is highly useful for limits involving powers. Ensure you correctly identify \(n\) and \(a\).

 

Question 17. Find the value of \( \lim_{x\rightarrow 1} \frac{\sqrt[3]{x}-1}{\sqrt{x}-1} \).
Answer: To find this limit, rewrite the roots as fractional exponents: \(\sqrt[3]{x} = x^{\frac{1}{3}}\) and \(\sqrt{x} = x^{\frac{1}{2}}\). Apply the standard limit formula \(\lim_{x \rightarrow a} \frac{x^n-a^n}{x-a} = n \cdot a^{n-1}\) to both the numerator and the denominator after dividing them by \((x-1)\). With \(a=1\), this simplifies to \(\frac{\frac{1}{3}(1)^{\frac{1}{3}-1}}{\frac{1}{2}(1)^{\frac{1}{2}-1}}\), which further reduces to \(\frac{1}{3} \div \frac{1}{2}\), giving the final value \(\frac{2}{3}\).
In simple words: Change the cube root and square root into fractions with powers. Then, use the special limit rule for powers on both the top and bottom of the main fraction. Since \(1\) raised to any power is \(1\), the problem becomes simple division of the power fractions.

🎯 Exam Tip: Convert root expressions to fractional exponents for easier application of power-rule limit formulas. Remember that \(\sqrt[n]{x} = x^{\frac{1}{n}}\).

Section E

I. Answer the following as required:

 

Question 1. If y = 5x + 7 then using tabular method, prove that when x → 2, y → 17.
Answer: To prove this using the tabular method, values of \(x\) that are very close to \(2\) are chosen, both slightly less than \(2\) (increasing towards \(2\)) and slightly greater than \(2\) (decreasing towards \(2\)). For each \(x\) value, the corresponding \(y\) value for the function \(y = 5x+7\) is calculated. The table below shows that as \(x\) gets closer to \(2\) from both sides, the calculated \(y\) values get closer to \(17\). This demonstrates that as \(x\) approaches \(2\), \(y\) approaches \(17\).

X increasing to 2	y = 5x + 7	X decreasing to 2	y
1.9	16.5	2.1	17.5
1.99	16.95	2.01	17.05
1.999	16.995	2.001	17.005
1.9999	16.9995	2.0001	17.0001
...	...	...	...

In simple words: To show this, pick numbers for \(x\) that are very close to \(2\). Then, calculate \(y\) for each \(x\). You will see that as \(x\) gets closer to \(2\), \(y\) gets closer to \(17\), proving the statement.

🎯 Exam Tip: When constructing a table for the tabular method, ensure you choose values that approach the limit from both sides (less than and greater than the limit value).

 

Question 2. If \( y = \frac{3 x^{2}+16 x+16}{x+4} \) then using tabular method, prove that when x → – 4, y → – 8.
Answer: First, simplify the given function \(y = \frac{3x^2+16x+16}{x+4}\) by factoring the numerator as \((x+4)(3x+4)\) and canceling out the common term \((x+4)\), which results in \(y = 3x+4\). Then, for the tabular method, select values of \(x\) very close to \(-4\), approaching from both smaller and larger sides. Calculate the corresponding \(y\) values for each \(x\). The table below demonstrates that as \(x\) approaches \(-4\), the \(y\) values get closer to \(-8\), thus proving the statement.

X increasing to -4	y = 3x + 4	X decreasing to -4	y
-4.1	-8.3	-3.9	-7.7
-4.01	-8.03	-3.99	-7.97
-4.001	-8.003	-3.999	-7.997
-4.0001	-8.0003	-3.9999	-7.9997
...	...	...	...

In simple words: First, simplify the given equation by factoring and canceling terms. Then, choose numbers for \(x\) that are very close to \(-4\) from both sides. Calculate \(y\) for each \(x\). You will see that as \(x\) gets closer to \(-4\), \(y\) gets closer to \(-8\), proving the statement.

🎯 Exam Tip: Simplifying the function algebraically before applying the tabular method can make calculations much easier and clearer.

 

Question 3. Using tabular method, prove that \( \lim_{x\rightarrow -1} \frac{3}{x+1} \) does not exist.
Answer: To show that the limit of \(f(x) = \frac{3}{x+1}\) does not exist as \(x\) approaches \(-1\) using a tabular method, one would choose \(x\) values very close to \(-1\) from both sides. When \(x\) approaches \(-1\) from the left side (e.g., -1.1, -1.01), \(f(x)\) would tend towards negative infinity. When \(x\) approaches \(-1\) from the right side (e.g., -0.9, -0.99), \(f(x)\) would tend towards positive infinity. Since \(f(x)\) does not approach a single value from both directions, the limit does not exist.
ℹ️ चित्र व्याख्या (Diagram Explanation): The source document indicates a table should be here, but the content for the table is missing from the provided text. To demonstrate the limit does not exist, a table would typically show \(x\) values approaching -1 from values smaller than -1 (e.g., -1.1, -1.01, -1.001) resulting in very large negative \(f(x)\) values, and \(x\) values approaching -1 from values larger than -1 (e.g., -0.9, -0.99, -0.999) resulting in very large positive \(f(x)\) values. This divergence shows no single limit.
In simple words: To prove this, pick numbers for \(x\) very close to \(-1\) from both sides. You would find that \(f(x)\) goes to a very big negative number from one side and a very big positive number from the other. Because it doesn't settle on one number, the limit does not exist.

🎯 Exam Tip: A limit does not exist if the function approaches different values (or infinities) from the left and right sides of the limit point.

II. Find the values of the following using tabular method:

 

Question 1. Find the value of \( \lim_{x\rightarrow 5} \frac{x^2-3x-10}{x-5} \) using the tabular method.
Answer: First, simplify the function \(f(x) = \frac{x^2-3x-10}{x-5}\). When \(x=5\), the function is in \(\frac{0}{0}\) form, meaning \((x-5)\) is a common factor. Factor the numerator as \((x-5)(x+2)\) and cancel \((x-5)\) to get \(f(x) = x+2\). To use the tabular method, choose values of \(x\) very close to \(5\) from both sides. Calculate \(f(x)\) for these values. The table below shows that as \(x\) approaches \(5\), \(f(x)\) approaches \(7\).

X increasing to 5	f(x) = x + 2	X decreasing to 5	f(x)
4.9	6.9	5.1	7.1
4.99	6.99	5.01	7.01
4.999	6.999	5.001	7.001
4.9999	6.9999	5.0001	7.0001
...	...	...	...

In simple words: First, simplify the fraction by factoring the top and canceling a matching part from the bottom. This leaves a simple equation. Then, use numbers very close to \(5\) for \(x\) and find the answer. You will see the answer gets close to \(7\).

🎯 Exam Tip: Always simplify the function algebraically first if substitution leads to an indeterminate form like \(\frac{0}{0}\). This simplifies the tabular calculations.

 

Question 2. Find the value of \( \lim_{x\rightarrow 1} \frac{2x^2+3x-5}{x-1} \) using the tabular method.
Answer: First, simplify the function \(f(x) = \frac{2x^2+3x-5}{x-1}\). When \(x=1\), the expression is in the indeterminate form \(\frac{0}{0}\). Factor the numerator as \((2x+5)(x-1)\) and cancel the common factor \((x-1)\) to get \(f(x) = 2x+5\). To use the tabular method, select \(x\) values very close to \(1\) from both sides and calculate their corresponding \(f(x)\) values. The table below shows that as \(x\) approaches \(1\), \(f(x)\) approaches \(7\).

X increasing to 1	f(x) = 2x + 5	X decreasing to 1	f(x)
0.9	6.8	1.1	7.2
0.99	6.98	1.01	7.02
0.999	6.998	1.001	7.002
0.9999	6.9998	1.0001	7.0002
...	...	...	...

In simple words: First, simplify the fraction by factoring the top and removing the common part from the bottom. This gives a simpler equation. Then, use numbers for \(x\) that are very close to \(1\) and find the answer. You will see the answer gets close to \(7\).

🎯 Exam Tip: Always show the algebraic simplification steps before presenting the tabular method, especially when dealing with indeterminate forms.

 

Question 3. Find the value of \( \lim_{x\rightarrow -1} \frac{4x^2+5x+1}{x+1} \) using the tabular method.
Answer: First, simplify the function \(f(x) = \frac{4x^2+5x+1}{x+1}\). Since substituting \(x=-1\) results in \(\frac{0}{0}\), factor the numerator as \((4x+1)(x+1)\) and cancel the common factor \((x+1)\), yielding \(f(x) = 4x+1\). To apply the tabular method, select values of \(x\) very close to \(-1\) from both sides. Calculate the corresponding \(f(x)\) values. The table below shows that as \(x\) approaches \(-1\), \(f(x)\) approaches \(-3\).

X increasing to -1	f(x) = 4x + 1	X decreasing to -1	f(x)
-1.1	-3.4	-0.9	-2.6
-1.01	-3.04	-0.99	-2.96
-1.001	-3.004	-0.999	-2.996
-1.0001	-3.0004	-0.9999	-2.9996
...	...	...	...

In simple words: First, make the fraction simpler by factoring the top and removing the common part from the bottom. This leaves a simple equation. Then, use numbers very close to \(-1\) for \(x\) and find the answer. You will see the answer gets close to \(-3\).

🎯 Exam Tip: Be careful with negative numbers during factorization and substitution. Double-check your arithmetic to avoid simple sign errors.

 

Question 4. Find the value of \( \lim_{x\rightarrow 0} 3x-1 \) using the tabular method.
Answer: To find the limit of \(f(x) = 3x-1\) as \(x\) approaches \(0\) using the tabular method, select values of \(x\) very close to \(0\) from both positive and negative sides. Calculate the corresponding \(f(x)\) values. The table below clearly shows that as \(x\) approaches \(0\), the values of \(f(x)\) approach \(-1\).

X increasing to 0	f(x) = 3x - 1	X decreasing to 0	f(x)
-0.1	-1.3	0.1	-0.7
-0.01	-1.03	0.01	-0.97
-0.001	-1.003	0.001	-0.997
-0.0001	-1.0003	0.0001	-0.9997
...	...	...	...

In simple words: To find this limit, pick numbers for \(x\) very close to \(0\) from both sides. Then, calculate the answer for \(f(x) = 3x-1\). You will see that as \(x\) gets closer to \(0\), the answer for \(f(x)\) gets closer to \(-1\).

🎯 Exam Tip: For simple polynomial functions, direct substitution is sufficient, but the tabular method confirms the limit intuitively by showing convergence from both sides.

III. Find The Values Of The Following:

 

Question 1. \( \lim_{h \to 0} \frac{(x + h)^7 - x^7}{h} \)

Answer: To find this limit, we first substitute \( t = x + h \).
As \( h \) approaches 0, \( t \) approaches \( x \). So, \( h = t - x \).
The expression becomes: \( \lim_{t \to x} \frac{t^7 - x^7}{t - x} \)
Using the standard limit formula \( \lim_{y \to a} \frac{y^n - a^n}{y - a} = n \cdot a^{n-1} \), we can solve this.
Here, \( n = 7 \) and \( a = x \).
So, the value of the limit is \( 7 \cdot x^{7-1} = 7x^6 \).
In simple words: This limit calculates the derivative of the function \( f(y) = y^7 \) with respect to \( y \) at the point \( y=x \). It shows how \( x^7 \) changes when \( x \) increases by a tiny amount \( h \).

🎯 Exam Tip: Recognizing and applying the standard limit formula \( \lim_{y \to a} \frac{y^n - a^n}{y - a} = n \cdot a^{n-1} \) is crucial for efficiently solving problems of this type. Careful substitution and identification of \( n \) and \( a \) prevent errors.

 

Question 2. \( \lim_{x \to 0} \frac{(1 + x)^{\frac{1}{10}} - 1}{x} \)

Answer: To evaluate this limit, we introduce a substitution: Let \( t = 1 + x \).
As \( x \) approaches 0, \( t \) approaches 1. Also, \( x = t - 1 \).
The limit expression transforms into: \( \lim_{t \to 1} \frac{t^{\frac{1}{10}} - 1}{t - 1} \)
We can rewrite 1 as \( 1^{\frac{1}{10}} \). So the expression is: \( \lim_{t \to 1} \frac{t^{\frac{1}{10}} - 1^{\frac{1}{10}}}{t - 1} \)
Using the standard limit formula \( \lim_{y \to a} \frac{y^n - a^n}{y - a} = n \cdot a^{n-1} \),
Here, \( n = \frac{1}{10} \) and \( a = 1 \).
Therefore, the value of the limit is \( \frac{1}{10} \cdot (1)^{\frac{1}{10}-1} = \frac{1}{10} \cdot (1)^{-\frac{9}{10}} = \frac{1}{10} \cdot 1 = \frac{1}{10} \).
In simple words: This limit finds the derivative of the function \( f(y) = y^{\frac{1}{10}} \) at \( y=1 \), or effectively, the derivative of \( (1+x)^{\frac{1}{10}} \) at \( x=0 \).

🎯 Exam Tip: When faced with expressions like \( ((1+x)^n - 1) / x \), consider the substitution method to align with the standard limit formula, as it's a direct application of the derivative definition at zero.

 

Question 3. \( \lim_{x \to 0} \frac{(1+x)^n - 1}{x} \)

Answer: To find the value of this limit, let \( t = 1 + x \).
When \( x \) approaches 0, \( t \) approaches 1. Also, \( x = t - 1 \).
The expression for the limit becomes: \( \lim_{t \to 1} \frac{t^n - 1}{t - 1} \)
We can write 1 as \( 1^n \). So the expression is: \( \lim_{t \to 1} \frac{t^n - 1^n}{t - 1} \)
Applying the standard limit formula \( \lim_{y \to a} \frac{y^n - a^n}{y - a} = n \cdot a^{n-1} \),
Here, \( n \) remains \( n \) and \( a = 1 \).
Thus, the limit is equal to \( n \cdot (1)^{n-1} = n \cdot 1 = n \).
In simple words: This limit calculates the derivative of \( f(y) = y^n \) at \( y=1 \), which is equivalent to finding the derivative of \( (1+x)^n \) at \( x=0 \).

🎯 Exam Tip: This is a fundamental limit related to the binomial expansion or the derivative of \( x^n \). Remembering that \( \lim_{x \to 0} \frac{(1+x)^n - 1}{x} = n \) can save significant time during exams.

 

Question 4. \( \lim_{x \to \frac{1}{2}} \frac{f(x) - f(\frac{1}{2})}{2x - 1} \) where \( f(x) = x^2 + x - 1 \)

Answer: First, we calculate \( f(\frac{1}{2}) \):
\( f(\frac{1}{2}) = (\frac{1}{2})^2 + \frac{1}{2} - 1 = \frac{1}{4} + \frac{1}{2} - 1 = \frac{1+2-4}{4} = -\frac{1}{4} \).
Now substitute \( f(x) \) and \( f(\frac{1}{2}) \) into the limit expression:
\( \lim_{x \to \frac{1}{2}} \frac{(x^2 + x - 1) - (-\frac{1}{4})}{2x - 1} \)
\( = \lim_{x \to \frac{1}{2}} \frac{x^2 + x - 1 + \frac{1}{4}}{2x - 1} \)
\( = \lim_{x \to \frac{1}{2}} \frac{x^2 + x - \frac{3}{4}}{2x - 1} \)
To simplify the numerator, multiply by 4 to get rid of the fraction:
\( = \lim_{x \to \frac{1}{2}} \frac{\frac{4x^2 + 4x - 3}{4}}{2x - 1} = \lim_{x \to \frac{1}{2}} \frac{4x^2 + 4x - 3}{4(2x - 1)} \)
Factor the numerator \( 4x^2 + 4x - 3 \):
We need two numbers that multiply to \( 4 \times (-3) = -12 \) and add to 4. These are 6 and -2.
\( 4x^2 + 6x - 2x - 3 = 2x(2x+3) - 1(2x+3) = (2x-1)(2x+3) \)
Substitute this back into the limit expression:
\( = \lim_{x \to \frac{1}{2}} \frac{(2x-1)(2x+3)}{4(2x - 1)} \)
Since \( x \to \frac{1}{2} \) and \( x \neq \frac{1}{2} \), we can cancel the common factor \( (2x - 1) \):
\( = \lim_{x \to \frac{1}{2}} \frac{2x+3}{4} \)
Now substitute \( x = \frac{1}{2} \) into the simplified expression:
\( = \frac{2(\frac{1}{2})+3}{4} = \frac{1+3}{4} = \frac{4}{4} = 1 \)
In simple words: This limit represents the slope of the function \( f(x) = x^2 + x - 1 \) at the specific point where \( x = \frac{1}{2} \). We find the function value at that point, simplify the expression by factoring, and then plug in the value of \( x \).

🎯 Exam Tip: For limits involving indeterminate forms like \( 0/0 \), algebraic techniques like factoring the numerator and denominator to cancel common factors are essential. Be precise with fractions and signs during calculations.

 

Question 5. \( \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \) where \( f(x) = x^3 \)

Answer: Given \( f(x) = x^3 \).
Then \( f(x + h) = (x + h)^3 \).
Expand \( (x + h)^3 \) using the binomial formula: \( x^3 + 3x^2h + 3xh^2 + h^3 \).
Substitute \( f(x + h) \) and \( f(x) \) into the limit expression:
\( \lim_{h \to 0} \frac{(x^3 + 3x^2h + 3xh^2 + h^3) - x^3}{h} \)
Simplify the numerator:
\( = \lim_{h \to 0} \frac{3x^2h + 3xh^2 + h^3}{h} \)
Factor out \( h \) from the numerator:
\( = \lim_{h \to 0} \frac{h(3x^2 + 3xh + h^2)}{h} \)
Since \( h \to 0 \) and \( h \neq 0 \), we can cancel \( h \) from the numerator and denominator:
\( = \lim_{h \to 0} (3x^2 + 3xh + h^2) \)
Now, substitute \( h = 0 \) into the simplified expression:
\( = 3x^2 + 3x(0) + (0)^2 = 3x^2 + 0 + 0 = 3x^2 \)
In simple words: This limit is the definition of the derivative of the function \( f(x) = x^3 \). It tells us the rate at which \( x^3 \) changes as \( x \) changes.

🎯 Exam Tip: This problem directly tests the understanding of the first-principle definition of a derivative. Accurate expansion of binomials and careful factorization of \( h \) are key to reaching the correct answer.

 

Question 6. \( \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \) where \( f(x) = x^7 \)

Answer: Given \( f(x) = x^7 \).
The limit expression is \( \lim_{h \to 0} \frac{(x + h)^7 - x^7}{h} \).
This expression matches the standard form of a derivative definition.
Alternatively, we can use a substitution similar to Question 1: Let \( t = x + h \).
As \( h \) approaches 0, \( t \) approaches \( x \). So, \( h = t - x \).
The limit transforms into: \( \lim_{t \to x} \frac{t^7 - x^7}{t - x} \)
Applying the standard limit formula \( \lim_{y \to a} \frac{y^n - a^n}{y - a} = n \cdot a^{n-1} \),
Here, \( n = 7 \) and \( a = x \).
Therefore, the value of the limit is \( 7 \cdot x^{7-1} = 7x^6 \).
In simple words: This limit finds the instantaneous rate of change (derivative) of the function \( f(x) = x^7 \). It shows how much \( x^7 \) changes for a very small change in \( x \).

🎯 Exam Tip: Recognizing this as the derivative definition is important. If the form \( \lim_{y \to a} \frac{y^n - a^n}{y - a} \) is clearly applicable after substitution, it provides a much faster solution than expanding \( (x+h)^7 \).

 

Question 7. \( \lim_{x \to 2} \frac{f(x) - f(2)}{x - 2} \) where \( f(x) = \sqrt{x + 7} \)

Answer: First, calculate \( f(2) \):
\( f(2) = \sqrt{2 + 7} = \sqrt{9} = 3 \).
Substitute \( f(x) \) and \( f(2) \) into the limit expression:
\( \lim_{x \to 2} \frac{\sqrt{x + 7} - 3}{x - 2} \)
To simplify, let \( t = x + 7 \).
When \( x \) approaches 2, \( t \) approaches \( 2 + 7 = 9 \). Also, \( x - 2 = (t - 7) - 2 = t - 9 \).
The limit expression becomes: \( \lim_{t \to 9} \frac{\sqrt{t} - 3}{t - 9} \)
We can rewrite this as: \( \lim_{t \to 9} \frac{t^{\frac{1}{2}} - 9^{\frac{1}{2}}}{t - 9} \)
Applying the standard limit formula \( \lim_{y \to a} \frac{y^n - a^n}{y - a} = n \cdot a^{n-1} \),
Here, \( n = \frac{1}{2} \) and \( a = 9 \).
So, the value of the limit is \( \frac{1}{2} \cdot (9)^{\frac{1}{2}-1} = \frac{1}{2} \cdot (9)^{-\frac{1}{2}} \)
\( = \frac{1}{2} \cdot \frac{1}{\sqrt{9}} = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6} \).
In simple words: This limit finds the slope of the tangent line to the function \( f(x) = \sqrt{x+7} \) at the point where \( x=2 \). It measures how quickly the square root function changes at that specific point.

🎯 Exam Tip: For limits involving square roots leading to indeterminate forms, substitution (as used here) or multiplying by the conjugate are effective methods. Both aim to transform the expression into a solvable form, often leveraging standard limit formulas.

 

Question 8. \( \lim_{h \to 0} \frac{f(2+h)-f(2)}{h} \) where \( f(x) = 2x^2 + 3 \)

Answer: First, calculate \( f(2) \):
\( f(2) = 2(2)^2 + 3 = 2(4) + 3 = 8 + 3 = 11 \).
Next, calculate \( f(2+h) \):
\( f(2+h) = 2(2+h)^2 + 3 \)
\( = 2(4 + 4h + h^2) + 3 \)
\( = 8 + 8h + 2h^2 + 3 \)
\( = 11 + 8h + 2h^2 \).
Substitute these into the limit expression:
\( \lim_{h \to 0} \frac{(11 + 8h + 2h^2) - 11}{h} \)
Simplify the numerator:
\( = \lim_{h \to 0} \frac{8h + 2h^2}{h} \)
Factor out \( h \) from the numerator:
\( = \lim_{h \to 0} \frac{h(8 + 2h)}{h} \)
Since \( h \to 0 \) and \( h \neq 0 \), we can cancel \( h \):
\( = \lim_{h \to 0} (8 + 2h) \)
Now substitute \( h = 0 \) into the simplified expression:
\( = 8 + 2(0) = 8 \).
In simple words: This limit calculates the rate of change of the function \( f(x) = 2x^2 + 3 \) exactly at the point \( x=2 \). It is the definition of the derivative of the function at that specific point.

🎯 Exam Tip: This problem is a direct application of the derivative definition. Pay close attention to algebraic expansions, especially squaring binomials, and ensure all terms are correctly combined and simplified before canceling the common factor \( h \).

 

Question 9. \( \lim_{x \to 0} \frac{f(2 + x) - f(2 - x)}{2x} \) where \( f(x) = x^2 \)

Answer: Given \( f(x) = x^2 \).
Calculate \( f(2+x) \): \( f(2+x) = (2+x)^2 = 4 + 4x + x^2 \).
Calculate \( f(2-x) \): \( f(2-x) = (2-x)^2 = 4 - 4x + x^2 \).
Substitute these into the limit expression:
\( \lim_{x \to 0} \frac{(4 + 4x + x^2) - (4 - 4x + x^2)}{2x} \)
Simplify the numerator:
\( = \lim_{x \to 0} \frac{4 + 4x + x^2 - 4 + 4x - x^2}{2x} \)
\( = \lim_{x \to 0} \frac{8x}{2x} \)
Since \( x \to 0 \) and \( x \neq 0 \), we can cancel \( x \) from numerator and denominator:
\( = \lim_{x \to 0} \frac{8}{2} \)
\( = \lim_{x \to 0} 4 \)
\( = 4 \).
In simple words: This limit measures the local rate of change of the function \( f(x) = x^2 \) around \( x=2 \). It compares the function's value slightly above \( 2 \) with its value slightly below \( 2 \).

🎯 Exam Tip: Be careful with signs when subtracting \( f(2-x) \) from \( f(2+x) \). A common mistake is not distributing the negative sign correctly to all terms in \( f(2-x) \).

 

Question 10. \( \lim_{x \to 2} \frac{f(x) - f(2)}{x - 2} \) where \( f(x) = x^2 + x \)

Answer: First, calculate \( f(2) \):
\( f(2) = (2)^2 + 2 = 4 + 2 = 6 \).
Substitute \( f(x) \) and \( f(2) \) into the limit expression:
\( \lim_{x \to 2} \frac{(x^2 + x) - 6}{x - 2} \)
Factor the quadratic numerator \( x^2 + x - 6 \). We need two numbers that multiply to -6 and add to 1. These are 3 and -2.
So, \( x^2 + x - 6 = (x + 3)(x - 2) \).
Substitute the factored form back into the limit expression:
\( = \lim_{x \to 2} \frac{(x + 3)(x - 2)}{x - 2} \)
Since \( x \to 2 \) and \( x \neq 2 \), we can cancel the common factor \( (x - 2) \):
\( = \lim_{x \to 2} (x + 3) \)
Now substitute \( x = 2 \) into the simplified expression:
\( = 2 + 3 = 5 \).
In simple words: This limit finds the slope of the tangent line to the function \( f(x) = x^2 + x \) at the specific point where \( x=2 \). It measures the instantaneous rate of change of the function at that point.

🎯 Exam Tip: When dealing with indeterminate forms \( 0/0 \) involving quadratic expressions, factoring the numerator is usually the most straightforward method. Always look for a common factor that matches the denominator's indeterminate part.