Samacheer Kalvi Class 11 Business Maths Solutions Chapter 5 Differential Calculus Exercise 5.2

Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 5 Differential Calculus Ex 5.2

Samacheer Kalvi 11th Business Maths Differential Calculus Ex 5.2 Text Book Back Questions and Answers

Question 1. Evaluate the following:
(i) \( \lim_{x \to 2} \frac{x^3+2}{x+1} \)
(ii) \( \lim_{x \to \infty} \frac{2x+5}{x^2+3x+9} \)
(iii) \( \lim_{n \to \infty} \frac{\Sigma n}{n^2} \)
(iv) \( \lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{5x} \)
(v) \( \lim_{x \to a} \frac{x^{\frac{5}{8}}-a^{\frac{5}{8}}}{x^{\frac{2}{3}}-a^{\frac{2}{3}}} \)
(vi) \( \lim_{x \to 0} \frac{\sin^2 3x}{x^2} \)
Answer:
(i) To evaluate \( \lim_{x \to 2} \frac{x^3+2}{x+1} \), we can directly substitute \( x=2 \) into the expression, as the denominator does not become zero.
\( \lim_{x \to 2} \frac{x^3+2}{x+1} = \frac{2^3+2}{2+1} \)
\( = \frac{8+2}{3} \)
\( = \frac{10}{3} \)

(ii) To evaluate \( \lim_{x \to \infty} \frac{2x+5}{x^2+3x+9} \), we divide both the numerator and the denominator by the highest power of \( x \) in the denominator, which is \( x^2 \).
\( \lim_{x \to \infty} \frac{2x+5}{x^2+3x+9} = \lim_{x \to \infty} \frac{\frac{2x}{x^2}+\frac{5}{x^2}}{\frac{x^2}{x^2}+\frac{3x}{x^2}+\frac{9}{x^2}} \)
\( = \lim_{x \to \infty} \frac{\frac{2}{x}+\frac{5}{x^2}}{1+\frac{3}{x}+\frac{9}{x^2}} \)
As \( x \to \infty \), terms like \( \frac{2}{x}, \frac{5}{x^2}, \frac{3}{x}, \frac{9}{x^2} \) all approach 0.
\( = \frac{0+0}{1+0+0} \)
\( = 0 \)

(iii) To evaluate \( \lim_{n \to \infty} \frac{\Sigma n}{n^2} \), we use the formula for the sum of the first \( n \) natural numbers, \( \Sigma n = \frac{n(n+1)}{2} \).
\( \lim_{n \to \infty} \frac{\Sigma n}{n^2} = \lim_{n \to \infty} \frac{\frac{n(n+1)}{2}}{n^2} \)
\( = \lim_{n \to \infty} \frac{n(n+1)}{2n^2} \)
\( = \lim_{n \to \infty} \frac{n^2+n}{2n^2} \)
Now, divide both numerator and denominator by \( n^2 \).
\( = \lim_{n \to \infty} \frac{\frac{n^2}{n^2}+\frac{n}{n^2}}{\frac{2n^2}{n^2}} \)
\( = \lim_{n \to \infty} \frac{1+\frac{1}{n}}{2} \)
As \( n \to \infty \), \( \frac{1}{n} \) approaches 0.
\( = \frac{1+0}{2} \)
\( = \frac{1}{2} \)

(iv) To evaluate \( \lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{5x} \), we multiply the numerator and denominator by the conjugate of the numerator, which is \( \sqrt{1+x}+\sqrt{1-x} \). This helps to remove the square roots from the numerator.
\( \lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{5x} \times \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} \)
\( = \lim_{x \to 0} \frac{(1+x)-(1-x)}{5x(\sqrt{1+x}+\sqrt{1-x})} \)
\( = \lim_{x \to 0} \frac{1+x-1+x}{5x(\sqrt{1+x}+\sqrt{1-x})} \)
\( = \lim_{x \to 0} \frac{2x}{5x(\sqrt{1+x}+\sqrt{1-x})} \)
We can cancel \( x \) from the numerator and denominator (since \( x \neq 0 \) as \( x \to 0 \)).
\( = \lim_{x \to 0} \frac{2}{5(\sqrt{1+x}+\sqrt{1-x})} \)
Now substitute \( x=0 \).
\( = \frac{2}{5(\sqrt{1+0}+\sqrt{1-0})} \)
\( = \frac{2}{5(1+1)} \)
\( = \frac{2}{5(2)} \)
\( = \frac{2}{10} \)
\( = \frac{1}{5} \)

(v) To evaluate \( \lim_{x \to a} \frac{x^{\frac{5}{8}}-a^{\frac{5}{8}}}{x^{\frac{2}{3}}-a^{\frac{2}{3}}} \), we use the standard limit formula \( \lim_{x \to a} \frac{x^n-a^n}{x-a} = na^{n-1} \). We divide both the numerator and the denominator by \( (x-a) \).
\( = \lim_{x \to a} \frac{\frac{x^{\frac{5}{8}}-a^{\frac{5}{8}}}{x-a}}{\frac{x^{\frac{2}{3}}-a^{\frac{2}{3}}}{x-a}} \)
Applying the formula to both parts:
Numerator: \( \frac{5}{8} a^{\frac{5}{8}-1} = \frac{5}{8} a^{-\frac{3}{8}} \)
Denominator: \( \frac{2}{3} a^{\frac{2}{3}-1} = \frac{2}{3} a^{-\frac{1}{3}} \)
So, the limit becomes:
\( = \frac{\frac{5}{8} a^{-\frac{3}{8}}}{\frac{2}{3} a^{-\frac{1}{3}}} \)
\( = \frac{5}{8} \times \frac{3}{2} \times a^{-\frac{3}{8}} \times a^{\frac{1}{3}} \)
\( = \frac{15}{16} a^{-\frac{3}{8}+\frac{1}{3}} \)
\( = \frac{15}{16} a^{\frac{-9+8}{24}} \)
\( = \frac{15}{16} a^{-\frac{1}{24}} \)

(vi) To evaluate \( \lim_{x \to 0} \frac{\sin^2 3x}{x^2} \), we can rewrite the expression and use the standard limit \( \lim_{x \to 0} \frac{\sin kx}{x} = k \).
\( \lim_{x \to 0} \frac{\sin^2 3x}{x^2} = \lim_{x \to 0} \left( \frac{\sin 3x}{x} \times \frac{\sin 3x}{x} \right) \)
To apply the formula, we multiply and divide each \( \frac{\sin 3x}{x} \) term by 3.
\( = \lim_{x \to 0} \left( \frac{\sin 3x}{3x} \times 3 \times \frac{\sin 3x}{3x} \times 3 \right) \)
\( = \lim_{x \to 0} \left( 3 \times \frac{\sin 3x}{3x} \right) \times \lim_{x \to 0} \left( 3 \times \frac{\sin 3x}{3x} \right) \)
We know that \( \lim_{x \to 0} \frac{\sin 3x}{3x} = 1 \).
\( = (3 \times 1) \times (3 \times 1) \)
\( = 3 \times 3 \)
\( = 9 \)
In simple words: For limits, sometimes you plug in the number directly. Other times, for big numbers or special functions like square roots or sine, you have to do a bit more work. This might mean dividing by the biggest 'x' power, multiplying by a special "conjugate" to clear roots, or using known rules for sine functions. Each type of limit has its own trick to solve it.

🎯 Exam Tip: Remember the basic limit formulas and techniques for different indeterminate forms. For limits at infinity, divide by the highest power of x; for square roots, multiply by the conjugate; and for trigonometric functions, use \( \lim_{x \to 0} \frac{\sin x}{x}=1 \).

Question 2. If \( \lim _{x \rightarrow a} \frac{x^{9}-a^{9}}{x-a}=\lim _{x \rightarrow 3}(x+6) \), find the value of a.
Answer:
We are given the equation: \( \lim_{x \to a} \frac{x^9-a^9}{x-a} = \lim_{x \to 3} (x+6) \)
For the left-hand side (LHS), we use the standard limit formula: \( \lim_{x \to a} \frac{x^n-a^n}{x-a} = na^{n-1} \). Here, \( n=9 \).
LHS \( = 9a^{9-1} = 9a^8 \)
For the right-hand side (RHS), we directly substitute \( x=3 \).
RHS \( = 3+6 = 9 \)
Now, we set LHS equal to RHS:
\( 9a^8 = 9 \)
Divide both sides by 9:
\( a^8 = 1 \)
To find \( a \), we take the eighth root of both sides. This means \( a \) can be either 1 or -1.
\( a = \pm 1 \)
In simple words: We used a special rule for limits on the left side to change it into a simpler form with 'a'. For the right side, we just put in the number 3. Then, we made both sides equal and solved for 'a'. The number 'a' can be either 1 or -1 because when you multiply either of them by itself eight times, you get 1.

🎯 Exam Tip: Clearly identify and apply the correct limit formula for expressions of the form \( \frac{x^n-a^n}{x-a} \). Remember that even powers often yield both positive and negative roots.

Question 3. If \( \lim _{x \rightarrow 2} \frac{x^{n}-2^{n}}{x-2}=448 \), then find the least positive integer n.
Answer:
We are given the equation: \( \lim_{x \to 2} \frac{x^n-2^n}{x-2} = 448 \)
Using the standard limit formula \( \lim_{x \to a} \frac{x^n-a^n}{x-a} = na^{n-1} \), where \( a=2 \).
So, the left side becomes: \( n \cdot 2^{n-1} \)
Therefore, we have the equation: \( n \cdot 2^{n-1} = 448 \)
Now, we need to find the value of \( n \) by prime factorizing 448.
\( 448 = 2 \times 224 \)
\( = 2 \times 2 \times 112 \)
\( = 2 \times 2 \times 2 \times 56 \)
\( = 2 \times 2 \times 2 \times 2 \times 28 \)
\( = 2 \times 2 \times 2 \times 2 \times 2 \times 14 \)
\( = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 7 \)
So, \( 448 = 7 \times 2^6 \)
Now we can compare \( n \cdot 2^{n-1} \) with \( 7 \cdot 2^6 \).
By comparing, we can see that \( n=7 \).
Let's check: If \( n=7 \), then \( n \cdot 2^{n-1} = 7 \cdot 2^{7-1} = 7 \cdot 2^6 \). This matches. Thus, the least positive integer \( n \) is 7.
In simple words: We used a special math rule for the limit part, which turned it into a simpler expression involving 'n'. Then we had to figure out what 'n' was by breaking down the number 448 into its basic building blocks (prime factors) to make it look similar to the expression we got from the limit. By matching the parts, we found that 'n' must be 7.

🎯 Exam Tip: When faced with equations like \( n \cdot a^{n-1} = K \), prime factorization of \( K \) is a key strategy to find \( n \) by matching powers and bases.

Question 4. If \( f(x) = \frac{x^{7}-128}{x^{5}-32} \), then find \( \lim _{x \rightarrow 2} f(x) \).
Answer:
We need to find \( \lim_{x \to 2} f(x) = \lim_{x \to 2} \frac{x^7-128}{x^5-32} \).
First, notice that \( 128 = 2^7 \) and \( 32 = 2^5 \). So the expression can be written as:
\( \lim_{x \to 2} \frac{x^7-2^7}{x^5-2^5} \)
To apply the standard limit formula \( \lim_{x \to a} \frac{x^n-a^n}{x-a} = na^{n-1} \), we can divide both the numerator and the denominator by \( (x-2) \).
\( = \lim_{x \to 2} \frac{\frac{x^7-2^7}{x-2}}{\frac{x^5-2^5}{x-2}} \)
Now, we apply the formula to both the numerator and the denominator separately:
Numerator limit: \( \lim_{x \to 2} \frac{x^7-2^7}{x-2} = 7 \cdot 2^{7-1} = 7 \cdot 2^6 \)
Denominator limit: \( \lim_{x \to 2} \frac{x^5-2^5}{x-2} = 5 \cdot 2^{5-1} = 5 \cdot 2^4 \)
So, the overall limit becomes:
\( = \frac{7 \cdot 2^6}{5 \cdot 2^4} \)
We can simplify the powers of 2:
\( = \frac{7}{5} \cdot 2^{6-4} \)
\( = \frac{7}{5} \cdot 2^2 \)
\( = \frac{7}{5} \cdot 4 \)
\( = \frac{28}{5} \)
In simple words: We wanted to find the limit of a fraction as 'x' got close to 2. We saw that the top and bottom parts of the fraction looked like a special form of limit rule if we divided both by (x-2). So we applied that rule to the top and the bottom separately. After solving each part and putting them back together, we found the final answer.

🎯 Exam Tip: When evaluating limits of fractions that are indeterminate (like 0/0), check if the expression fits the form \( \frac{x^n-a^n}{x-a} \) for both numerator and denominator. Dividing by \( (x-a) \) is a common and effective technique.

Question 5. Let \( f(x) = \frac{a x+b}{x+1} \), if \( \lim _{x \rightarrow 0} f(x)=2 \) and \( \lim _{x \rightarrow \infty} f(x) = 1 \), then show that \( f(-2) = 0 \).
Answer:
We are given the function \( f(x) = \frac{ax+b}{x+1} \). We need to find the values of \( a \) and \( b \) using the given limit conditions.

**Condition 1:** \( \lim_{x \to 0} f(x) = 2 \)
Substitute \( x=0 \) into the function (since the denominator is not zero):
\( \lim_{x \to 0} \frac{ax+b}{x+1} = \frac{a(0)+b}{0+1} \)
\( = \frac{b}{1} \)
\( = b \)
Since the limit is 2, we have \( b=2 \).

**Condition 2:** \( \lim_{x \to \infty} f(x) = 1 \)
For a rational function, as \( x \to \infty \), the limit is the ratio of the coefficients of the highest power of \( x \) in the numerator and denominator.
\( \lim_{x \to \infty} \frac{ax+b}{x+1} \)
Divide numerator and denominator by \( x \).
\( = \lim_{x \to \infty} \frac{\frac{ax}{x}+\frac{b}{x}}{\frac{x}{x}+\frac{1}{x}} \)
\( = \lim_{x \to \infty} \frac{a+\frac{b}{x}}{1+\frac{1}{x}} \)
As \( x \to \infty \), \( \frac{b}{x} \to 0 \) and \( \frac{1}{x} \to 0 \).
\( = \frac{a+0}{1+0} \)
\( = a \)
Since the limit is 1, we have \( a=1 \).

Now we have found \( a=1 \) and \( b=2 \). So, the function \( f(x) \) is:
\( f(x) = \frac{1x+2}{x+1} \)
\( f(x) = \frac{x+2}{x+1} \)

**Finally, we need to show that \( f(-2)=0 \):**
Substitute \( x=-2 \) into the function \( f(x) \):
\( f(-2) = \frac{-2+2}{-2+1} \)
\( = \frac{0}{-1} \)
\( = 0 \)
Thus, it is shown that \( f(-2)=0 \).
In simple words: We used the two given limits to find the missing numbers 'a' and 'b' in the function. The first limit helped us find 'b' by putting zero into the function. The second limit, as 'x' got very big, helped us find 'a' by looking at the main parts of the fraction. Once we knew 'a' and 'b', we put them back into the function. Then, we checked if plugging in -2 for 'x' gave us zero, which it did.

🎯 Exam Tip: When given limits for a rational function, remember that \( \lim_{x \to 0} \) often helps find the constant term (like b), and \( \lim_{x \to \infty} \) helps find the ratio of leading coefficients (like a).