GSEB Class 12 Statistics Solutions Chapter 4 Limit Exercise 4.2

Gseb Solutions Class 12 Statistics Part 2 Chapter 4 Limit Ex 4.2

Gujarat Board Textbook Solutions Class 12 Statistics Part 2 Chapter 4 Limit Ex 4.2

Question 1. Find the values of the following using tabular method:
(1) \( \lim_{x \to 1} 2x+1 \)
Answer: Here, the given function is \( f(x) = 2x + 1 \). To find the limit using the tabular method, we create a table with values of x that are very close to 1. The table shows how the function behaves as x approaches 1.

From the table, it is clear that as the value of x gets closer to 1, both by increasing and decreasing, the value of \( f(x) \) approaches 3. This means that as x tends to 1, \( f(x) \) tends to 3.
\( \implies \lim_{x \to 1} 2x + 1 = 3 \)
In simple words: When we check numbers for x very close to 1, both just below and just above 1, the answer from the function always gets closer and closer to 3. So, the limit is 3.

🎯 Exam Tip: Students should understand how to construct a table for the tabular method and observe the trend of function values as x approaches the given point.

Question 1. Find the values of the following using tabular method:
(2) \( \lim_{x \to 3} \frac{x^{2}-2x-3}{x-3} \)
Answer: Here, the function is \( f(x) = \frac{x^{2}-2x-3}{x-3} \).
We can simplify the function by factoring the numerator:
\( f(x) = \frac{x^{2}-3x+x-3}{x-3} \)
\( \implies f(x) = \frac{x(x-3)+1(x-3)}{x-3} \)
\( \implies f(x) = \frac{(x+1)(x-3)}{x-3} \)
\( \implies f(x) = x + 1 \) (for \( x \neq 3 \))

Now, we prepare a table by taking values of x very close to 3 for \( f(x) = x+1 \).

The table shows that as x approaches 3 (both from values smaller than 3 and larger than 3), the value of \( f(x) \) gets closer to 4. Therefore, as x tends to 3, \( f(x) \) tends to 4.
\( \implies \lim_{x \to 3} \frac{x^{2}-2x-3}{x-3} = 4 \)
In simple words: First, we made the fraction simpler to \( x+1 \). Then, by looking at numbers very near to 3 for x, the answer from \( x+1 \) always gets closer to 4. So, the limit is 4.

🎯 Exam Tip: Simplifying the function before applying the tabular method can make the calculations easier and clearer. Always check for common factors.

Question 1. Find the values of the following using tabular method:
(3) \( \lim_{x \to 2} \frac{2x^{2}+3x-14}{x-2} \)
Answer: Here, the function is \( f(x) = \frac{2x^{2}+3x-14}{x-2} \).
We can simplify the function by factoring the numerator:
\( f(x) = \frac{2x^{2}+7x-4x-14}{x-2} \)
\( \implies f(x) = \frac{x(2x+7)-2(2x+7)}{x-2} \)
\( \implies f(x) = \frac{(x-2)(2x+7)}{x-2} \)
\( \implies f(x) = 2x + 7 \) (for \( x \neq 2 \))

Now, we prepare a table by taking values of x very close to 2 for \( f(x) = 2x+7 \).

The table clearly shows that as x approaches 2 (from both sides), the value of \( f(x) \) gets closer to 11. This means that as x tends to 2, \( f(x) \) tends to 11.
\( \implies \lim_{x \to 2} \frac{2x^{2}+3x-14}{x-2} = 11 \)
In simple words: After simplifying the function to \( 2x+7 \), we observe that when x is very close to 2, the function's value gets very close to 11. So, the limit is 11.

🎯 Exam Tip: Always factorize the numerator and denominator if possible to simplify the function, especially when dealing with indeterminate forms (e.g., 0/0).

Question 1. Find the values of the following using tabular method:
(4) \( \lim_{x \to -3} \frac{2x^{2}+9x+9}{x+3} \)
Answer: Here, the function is \( f(x) = \frac{2x^{2}+9x+9}{x+3} \).
We can simplify the function by factoring the numerator:
\( f(x) = \frac{2x^{2}+6x+3x+9}{x+3} \)
\( \implies f(x) = \frac{2x(x+3)+3(x+3)}{x+3} \)
\( \implies f(x) = \frac{(x+3)(2x+3)}{x+3} \)
\( \implies f(x) = 2x + 3 \) (for \( x \neq -3 \))

Now, we prepare a table by taking values of x very close to -3 for \( f(x) = 2x+3 \).

From the table, it is observed that as x approaches -3 (from both sides), the value of \( f(x) \) gets closer to -3. This indicates that as x tends to -3, \( f(x) \) tends to -3.
\( \implies \lim_{x \to -3} \frac{2x^{2}+9x+9}{x+3} = -3 \)
In simple words: After simplifying the function to \( 2x+3 \), we can see that as x gets very close to -3, the function's value comes very close to -3. So, the limit is -3.

🎯 Exam Tip: When evaluating limits at a point where direct substitution leads to an indeterminate form, algebraic simplification (like factorization) is a crucial first step.

Question 1. Find the values of the following using tabular method:
(5) \( \lim_{x \to 2} x \)
Answer: Here, the function is \( f(x) = x \). To find the limit, we prepare a table by taking values of x very close to 2.

It is evident from the table that as x approaches 2 (from both sides), the value of \( f(x) \) also approaches 2. Therefore, as x tends to 2, \( f(x) \) tends to 2.
\( \implies \lim_{x \to 2} x = 2 \)
In simple words: For the function \( f(x) = x \), whatever value x gets close to, the function's value also gets close to that same value. So, if x goes to 2, the function goes to 2.

🎯 Exam Tip: For simple identity functions like \( f(x)=x \), the limit as x approaches a value 'a' is simply 'a'. This is a foundational concept for understanding limits.

Question 2. Using tabular method, show that \( \lim_{x \to 3} \frac{2}{x-3} \) does not exist.
Answer: Here, the function is \( f(x) = \frac{2}{x-3} \). To show that the limit does not exist using the tabular method, we examine the values of \( f(x) \) as x gets very close to 3.

It is clear from the table that as x approaches 3 from values less than 3 (x increasing to 3), \( f(x) \) becomes a very large negative number, tending towards \( -\infty \). As x approaches 3 from values greater than 3 (x decreasing to 3), \( f(x) \) becomes a very large positive number, tending towards \( +\infty \). Since \( f(x) \) does not approach a single definite value as x approaches 3, the limit of this function does not exist.
\( \implies \lim_{x \to 3} \frac{2}{x-3} \) does not exist.
In simple words: When x gets close to 3 from numbers smaller than 3, the answer goes to a huge negative number. When x gets close to 3 from numbers larger than 3, the answer goes to a huge positive number. Because it doesn't settle on one number, the limit does not exist.

🎯 Exam Tip: A limit only exists if the function approaches the same finite value from both the left and the right side of the point. If the values go to positive or negative infinity, or different finite values, the limit does not exist.

Question 3. If \( y = \frac{x^{2}+x-6}{x-2} \), show that as \( x \to 2 \), then \( y \to 5 \) using tabular method.
Answer: Here, the given function is \( y = \frac{x^{2}+x-6}{x-2} \).
First, we simplify the function by factoring the numerator:
\( y = \frac{x^{2}+3x-2x-6}{x-2} \)
\( \implies y = \frac{x(x+3)-2(x+3)}{x-2} \)
\( \implies y = \frac{(x-2)(x+3)}{x-2} \)
\( \implies y = x + 3 \) (for \( x \neq 2 \))

We need to show that as x approaches 2, y approaches 5. So, we will find \( \lim_{x \to 2} x + 3 \) using the tabular method.

Now, we prepare a table by taking values of x very close to 2 for \( y = x+3 \).

From the table, it is clear that as x gets closer to 2 (from both sides), the value of y approaches 5. This confirms that as x tends to 2, y tends to 5. Hence, it is proved that when \( x \to 2 \), then \( y \to 5 \).
In simple words: First, we simplified y to \( x+3 \). Then, by looking at numbers very close to 2 for x, the value of y always gets closer to 5. So, we have shown that as x goes to 2, y goes to 5.

🎯 Exam Tip: For "show that" questions involving limits, clearly state the simplified function and then systematically demonstrate the convergence using the tabular method.

Question 4. If \( y = 5 - 2x \), show that as \( x \to -1 \) then \( y \to 7 \) using tabular method.
Answer: Here, the given function is \( y = 5 - 2x \). We need to show that as x approaches -1, y approaches 7 using the tabular method.

We prepare a table by taking values of x very close to -1 for \( y = 5 - 2x \).

It is evident from the table that as x gets closer to -1 (from both sides), the value of y approaches 7. This confirms that as x tends to -1, y tends to 7. Hence, it is proved that when \( x \to -1 \), then \( y \to 7 \).
In simple words: When we pick numbers for x that are very close to -1 (both a little smaller and a little larger), the calculation for y always gives an answer very close to 7. This shows that when x approaches -1, y approaches 7.

🎯 Exam Tip: When evaluating limits for linear functions, direct substitution often works. The tabular method is used here for demonstration, but understanding direct substitution is also vital.