RBSE Solutions Class 11 Maths Chapter 10 Limits and Derivatives Exercise 10.2

Question 1. Write the repeating decimal for each of the following and use a bar to show the repetend.
(i) \( \lim_{x \to 2} \frac{x^2-3x+2}{x^2+x-6} \)
(ii) \( \lim_{x \to 1} \frac{(2x-3)(x-1)}{2x^2+x-3} \)
Answer:
(i) We need to evaluate the limit \( \lim_{x \to 2} \frac{x^2-3x+2}{x^2+x-6} \).
First, factor the numerator and the denominator:
\( x^2-3x+2 = x(x-2)-1(x-2) = (x-2)(x-1) \)
\( x^2+x-6 = x(x+3)-2(x+3) = (x+3)(x-2) \)
So, the limit becomes:
\( \lim_{x \to 2} \frac{(x-2)(x-1)}{(x+3)(x-2)} \)
Cancel out the common factor \( (x-2) \) (since \( x \ne 2 \) as \( x \to 2 \)):
\( \lim_{x \to 2} \frac{x-1}{x+3} \)
Now, substitute \( x=2 \):
\( \frac{2-1}{2+3} = \frac{1}{5} \)
The value of the limit is \( \frac{1}{5} \).
(ii) We need to evaluate the limit \( \lim_{x \to 1} \frac{(2x-3)(x-1)}{2x^2+x-3} \).
First, expand and factor the denominator:
\( 2x^2+x-3 = 2x^2+3x-2x-3 = x(2x+3)-1(2x+3) = (2x+3)(x-1) \)
So, the limit becomes:
\( \lim_{x \to 1} \frac{(2x-3)(x-1)}{(2x+3)(x-1)} \)
Cancel out the common factor \( (x-1) \) (since \( x \ne 1 \) as \( x \to 1 \)):
\( \lim_{x \to 1} \frac{2x-3}{2x+3} \)
Now, substitute \( x=1 \):
\( \frac{2(1)-3}{2(1)+3} = \frac{2-3}{2+3} = \frac{-1}{5} \)
The value of the limit is \( \frac{-1}{5} \).
In simple words: For limits, first try to simplify the expression by factoring. If a term in the denominator becomes zero when you substitute the limit value, try to cancel it out. After simplifying, put in the number \( x \) is approaching to get the final answer.

🎯 Exam Tip: Always try to factorize the numerator and denominator to eliminate any terms that make the expression undefined at the limit point. This is crucial for solving indeterminate forms like \( \frac{0}{0} \).

Question 2. Solve the following limits.
(i) \( \lim_{\alpha \to \pi/4} \frac{\sin \alpha - \cos \alpha}{\alpha-\pi/4} \)
(ii) \( \lim_{x \to 1} \frac{x^n-1}{x-1} \)
Answer:
(i) We need to evaluate the limit \( \lim_{\alpha \to \pi/4} \frac{\sin \alpha - \cos \alpha}{\alpha-\pi/4} \).
We know that \( \sin \alpha - \cos \alpha = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin \alpha - \frac{1}{\sqrt{2}} \cos \alpha \right) \)
\( = \sqrt{2} \left( \cos(\pi/4) \sin \alpha - \sin(\pi/4) \cos \alpha \right) \)
\( = \sqrt{2} \sin(\alpha - \pi/4) \)
Let \( \theta = \alpha - \pi/4 \). As \( \alpha \to \pi/4 \), \( \theta \to 0 \).
The limit becomes:
\( \lim_{\theta \to 0} \frac{\sqrt{2} \sin \theta}{\theta} \)
We know that \( \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \).
So, \( \sqrt{2} \times 1 = \sqrt{2} \).
The value of the limit is \( \sqrt{2} \).
(ii) We need to evaluate the limit \( \lim_{x \to 1} \frac{x^n-1}{x-1} \).
Let \( x-1=y \). So, \( x = 1+y \). As \( x \to 1 \), \( y \to 0 \).
The expression \( x^n-1 = (1+y)^n-1 \).
Using the binomial expansion \( (1+y)^n = 1 + ny + \frac{n(n-1)}{2!}y^2 + \frac{n(n-1)(n-2)}{3!}y^3 + \dots \)
So, \( (1+y)^n-1 = ny + \frac{n(n-1)}{2!}y^2 + \frac{n(n-1)(n-2)}{3!}y^3 + \dots \)
The limit becomes:
\( \lim_{y \to 0} \frac{ny + \frac{n(n-1)}{2!}y^2 + \frac{n(n-1)(n-2)}{3!}y^3 + \dots}{y} \)
Divide each term in the numerator by \( y \):
\( \lim_{y \to 0} \left( n + \frac{n(n-1)}{2!}y + \frac{n(n-1)(n-2)}{3!}y^2 + \dots \right) \)
Now, substitute \( y=0 \):
\( n + \frac{n(n-1)}{2!}(0) + \frac{n(n-1)(n-2)}{3!}(0)^2 + \dots = n+0+0+\dots = n \)
The value of the limit is \( n \).
In simple words: For limits involving trigonometric functions, sometimes you need to change the form of the expression using identities. For expressions with powers, a substitution can turn it into a binomial expansion. Remember the special limit \( \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \) and the standard derivative definition form for \( x^n \).

🎯 Exam Tip: For problems like Q2(i), remember to use trigonometric identities to transform the expression into a known limit form. For Q2(ii), recognize this as the definition of the derivative of \( x^n \) at \( x=1 \), which is \( n \cdot 1^{n-1} = n \). This can save time if L'Hopital's Rule is allowed.

Question 3. Evaluate the following limits.
(i) \( \lim_{x \to 0} \frac{2^x-1}{\sqrt{1+x}-1} \)
(ii) \( \lim_{x \to 0} \frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{x^2} \)
Answer:
(i) We need to evaluate the limit \( \lim_{x \to 0} \frac{2^x-1}{\sqrt{1+x}-1} \).
We can use the series expansions for \( 2^x \) and \( \sqrt{1+x} \).
\( 2^x = e^{x \log 2} = 1 + x \log 2 + \frac{(x \log 2)^2}{2!} + \frac{(x \log 2)^3}{3!} + \dots \)
So, \( 2^x-1 = x \log 2 + \frac{x^2 (\log 2)^2}{2!} + \frac{x^3 (\log 2)^3}{3!} + \dots \)
And \( \sqrt{1+x} = (1+x)^{1/2} = 1 + \frac{1}{2}x + \frac{(1/2)(1/2-1)}{2!}x^2 + \dots = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \dots \)
So, \( \sqrt{1+x}-1 = \frac{1}{2}x - \frac{1}{8}x^2 + \dots \)
The limit becomes:
\( \lim_{x \to 0} \frac{x \log 2 + \frac{x^2 (\log 2)^2}{2!} + \frac{x^3 (\log 2)^3}{3!} + \dots}{\frac{1}{2}x - \frac{1}{8}x^2 + \dots} \)
Divide both numerator and denominator by \( x \):
\( \lim_{x \to 0} \frac{\log 2 + \frac{x (\log 2)^2}{2!} + \frac{x^2 (\log 2)^3}{3!} + \dots}{\frac{1}{2} - \frac{1}{8}x + \dots} \)
Now, substitute \( x=0 \):
\( \frac{\log 2 + 0 + 0 + \dots}{\frac{1}{2} - 0 + 0 + \dots} = \frac{\log 2}{1/2} = 2 \log 2 \)
The value of the limit is \( 2 \log 2 \).
(ii) We need to evaluate the limit \( \lim_{x \to 0} \frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{x^2} \).
Multiply the numerator and denominator by the conjugate of the numerator, which is \( \sqrt{1+x^2}+\sqrt{1-x^2} \):
\( \lim_{x \to 0} \frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{x^2} \times \frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}} \)
\( = \lim_{x \to 0} \frac{(1+x^2)-(1-x^2)}{x^2(\sqrt{1+x^2}+\sqrt{1-x^2})} \)
\( = \lim_{x \to 0} \frac{1+x^2-1+x^2}{x^2(\sqrt{1+x^2}+\sqrt{1-x^2})} \)
\( = \lim_{x \to 0} \frac{2x^2}{x^2(\sqrt{1+x^2}+\sqrt{1-x^2})} \)
Cancel out \( x^2 \) (since \( x \ne 0 \) as \( x \to 0 \)):
\( = \lim_{x \to 0} \frac{2}{\sqrt{1+x^2}+\sqrt{1-x^2}} \)
Now, substitute \( x=0 \):
\( = \frac{2}{\sqrt{1+0}+\sqrt{1-0}} = \frac{2}{1+1} = \frac{2}{2} = 1 \)
The value of the limit is \( 1 \).
In simple words: For limits that look like \( \frac{0}{0} \) directly, you can use known series expansions for functions like \( 2^x \) or \( \sqrt{1+x} \). For expressions with square roots in the numerator, multiplying by the conjugate helps simplify the expression by removing the roots and often reveals a common factor that can be cancelled.

🎯 Exam Tip: When faced with limits involving complex functions like \( a^x \) or \( (1+x)^n \), consider using Maclaurin series expansions to simplify the terms. For square root differences, rationalization is a powerful technique.

Question 4. (i) \( \lim_{x \to 0} \frac{e^x+e^{-x}-2 \cos x}{x \sin x} \)
Answer:
We need to evaluate the limit \( \lim_{x \to 0} \frac{e^x+e^{-x}-2 \cos x}{x \sin x} \).
We use the series expansions for \( e^x \), \( e^{-x} \), \( \cos x \), and \( \sin x \) around \( x=0 \):
\( e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\dots \)
\( e^{-x} = 1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}-\dots \)
\( \cos x = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\dots \)
\( \sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots \)
Now substitute these into the numerator:
\( e^x+e^{-x} = (1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots) + (1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\dots) = 2 + 2\frac{x^2}{2!} + 2\frac{x^4}{4!} + \dots \)
\( 2 \cos x = 2(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\dots) = 2 - 2\frac{x^2}{2!} + 2\frac{x^4}{4!} - \dots \)
Numerator: \( (e^x+e^{-x}) - 2 \cos x = (2 + x^2 + \frac{x^4}{12} + \dots) - (2 - x^2 + \frac{x^4}{12} - \dots) \)
\( = 2x^2 + \text{higher powers of } x \)
Now, the denominator \( x \sin x = x(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots) = x^2-\frac{x^4}{3!}+\frac{x^6}{5!}-\dots \)
The limit becomes:
\( \lim_{x \to 0} \frac{2x^2 + \text{higher powers of } x}{x^2-\frac{x^4}{6}+\dots} \)
Divide both numerator and denominator by \( x^2 \):
\( \lim_{x \to 0} \frac{2 + \text{terms with } x}{\frac{x^2}{1} - \frac{x^4}{6}+\dots} = \lim_{x \to 0} \frac{2 + \dots}{1-\frac{x^2}{6}+\dots} \)
Now, substitute \( x=0 \):
\( \frac{2}{1} = 2 \)
The value of the limit is \( 2 \).
(ii) We need to evaluate the limit \( \lim_{x \to 0} \frac{\sin^{-1} x - \tan^{-1} x}{x^3} \)
We use the series expansions for \( \sin^{-1} x \) and \( \tan^{-1} x \) around \( x=0 \):
\( \sin^{-1} x = x + \frac{1^2}{3!}x^3 + \frac{1^2 \cdot 3^2}{5!}x^5 + \dots = x + \frac{x^3}{6} + \frac{9x^5}{120} + \dots \)
\( \tan^{-1} x = x - \frac{x^3}{3} + \frac{x^5}{5} - \dots \)
Numerator: \( \sin^{-1} x - \tan^{-1} x = \left( x + \frac{x^3}{6} + \frac{x^5}{8} + \dots \right) - \left( x - \frac{x^3}{3} + \frac{x^5}{5} - \dots \right) \)
\( = (\frac{1}{6} + \frac{1}{3})x^3 + (\frac{1}{8} - \frac{1}{5})x^5 + \dots \)
\( = (\frac{1+2}{6})x^3 + (\frac{5-8}{40})x^5 + \dots \)
\( = \frac{3}{6}x^3 - \frac{3}{40}x^5 + \dots = \frac{1}{2}x^3 - \frac{3}{40}x^5 + \dots \)
The limit becomes:
\( \lim_{x \to 0} \frac{\frac{1}{2}x^3 - \frac{3}{40}x^5 + \dots}{x^3} \)
Divide by \( x^3 \):
\( \lim_{x \to 0} \left( \frac{1}{2} - \frac{3}{40}x^2 + \dots \right) \)
Now, substitute \( x=0 \):
\( \frac{1}{2} \)
The value of the limit is \( \frac{1}{2} \).
In simple words: When dealing with limits where substituting \( x=0 \) gives \( \frac{0}{0} \), using known series expansions for functions like \( e^x \), \( \cos x \), \( \sin^{-1} x \), and \( \tan^{-1} x \) helps simplify the expression. After expanding, cancel out the lowest power of \( x \) from both the top and bottom, then substitute \( x=0 \) to find the limit.

🎯 Exam Tip: Remember the Maclaurin series for common functions: \( e^x \), \( \sin x \), \( \cos x \), \( \sin^{-1} x \), and \( \tan^{-1} x \). These are very useful for evaluating limits of indeterminate forms when L'Hopital's Rule is not permitted or is too cumbersome.

Question 5. Evaluate the following limits.
(i) \( \lim_{x \to \infty} \frac{x + \sin x}{x-\cos x} \)
(ii) \( \lim_{n \to \infty} \frac{\sqrt{3n^2-1}-\sqrt{2n^2-1}}{4n+3} \)
Answer:
(i) We need to evaluate the limit \( \lim_{x \to \infty} \frac{x + \sin x}{x-\cos x} \).
Divide both the numerator and the denominator by \( x \), the highest power of \( x \) in the expression:
\( \lim_{x \to \infty} \frac{1 + \frac{\sin x}{x}}{1-\frac{\cos x}{x}} \)
We know that as \( x \to \infty \), \( -1 \le \sin x \le 1 \) and \( -1 \le \cos x \le 1 \).
Therefore, \( \lim_{x \to \infty} \frac{\sin x}{x} = 0 \) and \( \lim_{x \to \infty} \frac{\cos x}{x} = 0 \).
Substitute these values into the limit expression:
\( \frac{1+0}{1-0} = 1 \)
The value of the limit is \( 1 \).
(ii) We need to evaluate the limit \( \lim_{n \to \infty} \frac{\sqrt{3n^2-1}-\sqrt{2n^2-1}}{4n+3} \).
Divide both the numerator and the denominator by \( n \), the highest power of \( n \) in the expression. For terms inside square roots, divide by \( n^2 \):
\( \lim_{n \to \infty} \frac{\sqrt{\frac{3n^2}{n^2}-\frac{1}{n^2}}-\sqrt{\frac{2n^2}{n^2}-\frac{1}{n^2}}}{\frac{4n}{n}+\frac{3}{n}} \)
\( = \lim_{n \to \infty} \frac{\sqrt{3-\frac{1}{n^2}}-\sqrt{2-\frac{1}{n^2}}}{4+\frac{3}{n}} \)
As \( n \to \infty \), \( \frac{1}{n^2} \to 0 \) and \( \frac{3}{n} \to 0 \).
Substitute these values into the limit expression:
\( = \frac{\sqrt{3-0}-\sqrt{2-0}}{4+0} \)
\( = \frac{\sqrt{3}-\sqrt{2}}{4} \)
The value of the limit is \( \frac{\sqrt{3}-\sqrt{2}}{4} \).
In simple words: For limits as \( x \) (or \( n \)) goes to infinity, divide every term by the highest power of \( x \) (or \( n \)) present. Remember that terms like \( \frac{\sin x}{x} \) or \( \frac{1}{x} \) will go to zero as \( x \) gets very large. Simplify and then substitute infinity.

🎯 Exam Tip: When dealing with limits at infinity, the key is to identify the highest power of the variable in the numerator and denominator. Divide all terms by this power. Remember that \( \lim_{x \to \infty} \frac{\text{bounded function}}{x^k} = 0 \) for \( k>0 \), which is useful for trigonometric functions.

Question 6. Evaluate the following limits.
(i) \( \lim_{x \to 0} \frac{\sin 5x-\sin 3x}{\sin x} \)
(ii) \( \lim_{x \to 0} \frac{e^{\sin x}-1}{x} \)
Answer:
(i) We need to evaluate the limit \( \lim_{x \to 0} \frac{\sin 5x-\sin 3x}{\sin x} \).
Divide both numerator and denominator by \( x \). We will use the standard limit \( \lim_{y \to 0} \frac{\sin y}{y} = 1 \).
\( = \lim_{x \to 0} \frac{\frac{\sin 5x}{x}-\frac{\sin 3x}{x}}{\frac{\sin x}{x}} \)
To apply the limit formula, multiply and divide by the coefficients of \( x \):
\( = \lim_{x \to 0} \frac{5 \cdot \frac{\sin 5x}{5x}-3 \cdot \frac{\sin 3x}{3x}}{1 \cdot \frac{\sin x}{x}} \)
Now apply the limit. As \( x \to 0 \), \( 5x \to 0 \) and \( 3x \to 0 \).
\( = \frac{5(1)-3(1)}{1} = \frac{5-3}{1} = 2 \)
The value of the limit is \( 2 \).
(ii) We need to evaluate the limit \( \lim_{x \to 0} \frac{e^{\sin x}-1}{x} \).
We know the standard limit formula \( \lim_{y \to 0} \frac{e^y-1}{y} = 1 \).
Let \( y = \sin x \). As \( x \to 0 \), \( \sin x \to 0 \), so \( y \to 0 \).
We can rewrite the limit as:
\( \lim_{x \to 0} \frac{e^{\sin x}-1}{\sin x} \cdot \frac{\sin x}{x} \)
This can be separated into two limits:
\( \left( \lim_{x \to 0} \frac{e^{\sin x}-1}{\sin x} \right) \cdot \left( \lim_{x \to 0} \frac{\sin x}{x} \right) \)
For the first part, let \( y = \sin x \). As \( x \to 0 \), \( y \to 0 \). So \( \lim_{y \to 0} \frac{e^y-1}{y} = 1 \).
For the second part, \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \).
Therefore, the overall limit is \( 1 \times 1 = 1 \).
The value of the limit is \( 1 \).
In simple words: When dealing with limits involving sine functions, use the property \( \lim_{y \to 0} \frac{\sin y}{y} = 1 \). For exponential limits like \( e^y-1 \), use the property \( \lim_{y \to 0} \frac{e^y-1}{y} = 1 \). Sometimes you need to adjust the expression by multiplying and dividing by terms to match these standard forms.

🎯 Exam Tip: Remember standard trigonometric and exponential limits. For \( \lim_{x \to 0} \frac{\sin(ax)}{\sin(bx)} = \frac{a}{b} \). For \( \lim_{x \to 0} \frac{e^{f(x)}-1}{x} \) where \( \lim_{x \to 0} f(x) = 0 \), it often simplifies to \( \lim_{x \to 0} \frac{f(x)}{x} \).

Question 7. Evaluate the following limits.
(i) \( \lim_{x \to 0} \frac{x(e^x-1)}{1-\cos x} \)
(ii) \( \lim_{x \to a} \frac{(x+2)^{3/2}-(a+2)^{3/2}}{x-a} \)
Answer:
(i) We need to evaluate the limit \( \lim_{x \to 0} \frac{x(e^x-1)}{1-\cos x} \).
We know the standard limits:
\( \lim_{x \to 0} \frac{e^x-1}{x} = 1 \)
\( \lim_{x \to 0} \frac{1-\cos x}{x^2} = \frac{1}{2} \)
Rewrite the given limit by multiplying and dividing by appropriate powers of \( x \):
\( \lim_{x \to 0} \frac{x \cdot (e^x-1)}{1-\cos x} = \lim_{x \to 0} \frac{x \cdot (e^x-1)}{x^2 \cdot \frac{1-\cos x}{x^2}} \)
\( = \lim_{x \to 0} \frac{\frac{e^x-1}{x}}{\frac{1-\cos x}{x^2}} \)
Now apply the limits to the numerator and denominator separately:
\( = \frac{\lim_{x \to 0} \frac{e^x-1}{x}}{\lim_{x \to 0} \frac{1-\cos x}{x^2}} = \frac{1}{1/2} = 2 \)
The value of the limit is \( 2 \).
(ii) We need to evaluate the limit \( \lim_{x \to a} \frac{(x+2)^{3/2}-(a+2)^{3/2}}{x-a} \).
We use the standard limit formula: \( \lim_{y \to c} \frac{y^n-c^n}{y-c} = nc^{n-1} \).
Let \( y = x+2 \) and \( c = a+2 \). As \( x \to a \), \( x+2 \to a+2 \), so \( y \to c \).
Also, \( x-a = (x+2)-(a+2) = y-c \).
Substitute these into the limit expression:
\( \lim_{y \to c} \frac{y^{3/2}-c^{3/2}}{y-c} \)
Using the formula with \( n=3/2 \):
\( = \frac{3}{2} c^{(3/2)-1} = \frac{3}{2} c^{1/2} \)
Substitute back \( c = a+2 \):
\( = \frac{3}{2} (a+2)^{1/2} \)
The value of the limit is \( \frac{3}{2} \sqrt{a+2} \).
In simple words: For limits involving exponential and cosine functions, rewrite the expression using known standard limit forms by dividing by appropriate powers of \( x \). For limits of the form \( \frac{y^n-c^n}{y-c} \), use the direct formula \( nc^{n-1} \) after making a suitable substitution to match the variables.

🎯 Exam Tip: Memorize the standard limits involving \( e^x-1 \), \( \sin x \), \( \tan x \), and \( 1-\cos x \) as \( x \to 0 \). Also, know the general form for \( \lim_{x \to a} \frac{x^n-a^n}{x-a} = na^{n-1} \) as it often appears in variations. These formulas are fundamental for evaluating such limits quickly and accurately.