Chapter 13: Limits and Derivatives

Ans: $f(x)= \left\{\begin{align} &  x + 2, x \leq -1 \\ &  c{x^2}, x >  - 1 \end{align} \right.$

${\text{LHL}} = \mathop {\lim }\limits_{x \to  - {1^ - }} f(x)$

$= \mathop {\lim }\limits_{x \to  - {1^ - }} (x + 2)$

$= \mathop {\lim }\limits_{x \to  - {1^ - }} (x + 2)$

$= \mathop {\lim }\limits_{h \to 0} (1 - h) = 1$

${\text{RHL}} = \mathop {\lim }\limits_{x \to  - {1^ + }} f(x)$

$= \mathop {\lim }\limits_{x \to  - {1^ + }} c{x^2}$

$= \mathop {\lim }\limits_{h \to 0} c{( - 1 + h)^2}$

$= c$

If $\mathop {\lim }\limits_{x \to  - 1} f\left( x \right)$ exist, then $LHL = RHL$

$\therefore c = 1$

Choose the correct answer out of 4 options given against each Exercise 54 to 76 (M.C.Q).

Ans: 

Given: 

$\mathop {\lim }\limits_{x \to \pi } \dfrac{{\sin x}}{{x - \pi }}$

We know that,

$\because \sin \left( {\pi  - \,\theta } \right)\, = \,\sin \theta$

$\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1{\text{ and }}\pi  - x \to 0 \Rightarrow x \to \pi$

Let,     

$= \mathop {\lim }\limits_{x \to \pi } \dfrac{{\sin x}}{{x - \pi }}$

$= \mathop {\lim }\limits_{x \to \pi } \dfrac{{\sin \left( {\pi  - x} \right)}}{{ - \left( {\pi  - x} \right)}}{\text{  }}\left[ {{\text{ }}\because \sin \left( {\pi  - \theta } \right) = \sin \theta {\text{ }}} \right]$

$=  - \mathop {\lim }\limits_{x \to \pi } \dfrac{{\sin \left( {\pi  - x} \right)}}{{\left( {\pi  - x} \right)}}\left[ {{\text{ }}\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1{\text{ and }}\pi  - x \to 0 \Rightarrow x \to \pi {\text{ }}} \right]{\text{ }}$

$=  - 1$ 

Correct Option:  C 

Ans: Given: 

$\mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2}\cos x}}{{1 - \cos x}}$

We know that,

$\because {\text{ }}1 - \cos x = 2{\sin ^2}\dfrac{x}{2}$

Let,

$\mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2}\cos x}}{{1 - \cos x}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2}\cos x}}{{2{{\sin }^2}\dfrac{x}{2}}}\quad \left[ {\because 1 - \cos x = 2{{\sin }^2}\dfrac{x}{2}} \right]$

$= 2\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\left( {\dfrac{x}{2}} \right)}^2}}}{{{{\sin }^2}\dfrac{x}{2}}} \cdot \mathop {\lim }\limits_{x \to 0} \cos x$

$= 2 \cdot 1 = 2$ 

Correct Option:  A

Ans: Given:

$\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\left( {1 + x} \right)}^n} - 1}}{x}$

We know that,

$\because \mathop {\lim }\limits_{x \to a} \dfrac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}$

Then,

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{{{\left( {1 + x} \right)}^n} - 1}}{x}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{{{(1 + x)}^n} - 1}}{{(1 + x) - 1}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{{{(1 + x)}^n} - {1^n}}}{{(1 + x) - 1}}$

$= \mathop {\lim }\limits_{(1 + x) \to 1} \dfrac{{{{(1 + x)}^n} - {1^n}}}{{(1 + x) - 1}}$

$= n \cdot {(1)^{n - 1}}$\[\left[ {\because \mathop {\lim }\limits_{x \to a} \dfrac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}} \right]\]

$= n$

Correct Option:  A

Ans: Given:

$\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^m} - 1}}{{{x^n} - 1}}$

We know that,

\[\because \mathop {\lim }\limits_{x \to a} \dfrac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}\]

Then,

$=\lim _{x \rightarrow 1 \atop \rightarrow 1} \dfrac{x^{m}-1}{x^{n}-1}$

$= \mathop {\lim }\limits_{x \to 1} \dfrac{{\dfrac{{{x^m} - 1}}{{x - 1}}}}{{\dfrac{{{x^n} - 1}}{{x - 1}}}}$ 

$= \dfrac{{\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^m} - {1^m}}}{{x - 1}}}}{{\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^n} - {1^n}}}{{x - 1}}}}$

$=\lim _{x \rightarrow 1 \atop \rightarrow 1} \dfrac{x^{m}-1}{x^{n}-1}$

$= \mathop {\lim }\limits_{x \to 1} \dfrac{{\dfrac{{{x^m} - 1}}{{x - 1}}}}{{\dfrac{{{x^n} - 1}}{{x - 1}}}}$

$= \dfrac{{\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^m} - {1^m}}}{{x - 1}}}}{{\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^n} - {1^n}}}{{x - 1}}}}$

$= \dfrac{{m{{(1)}^{m - 1}}}}{{n{{(1)}^{n - 1}}}}$

$= \dfrac{m}{n}\quad \left[ {\because \mathop {\lim }\limits_{x \to a} \dfrac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}} \right]$ 

Correct Option:  B

Ans: Given:

$\mathop {\lim }\limits_{\theta  \to 0} \dfrac{{1 - \cos 4\theta }}{{1 - \cos 6\theta }}$

We know that,

$\because 1 - \cos 2\theta  = 2{\sin ^2}\theta$

\[\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1{\text{ and }}x \to 0 \Rightarrow kx \to 0\]

Let,

$\mathop {\lim }\limits_{\theta  \to 0} \dfrac{{1 - \cos 4\theta }}{{1 - \cos 6\theta }}$

$= \mathop {\lim }\limits_{\theta  \to 0} \dfrac{{2{{\sin }^2}2\theta }}{{2{{\sin }^2}3\theta }}\quad \left[ {\because 1 - \cos 2\theta  = 2{{\sin }^2}\theta } \right]$

\[ = \dfrac{{\mathop {\lim }\limits_{\theta  \to 0} \dfrac{{{{\sin }^2}2\theta }}{{{{(2\theta )}^2}}} \cdot {{(2\theta )}^2}}}{{\mathop {\lim }\limits_{\theta  \to 0} \dfrac{{{{\sin }^2}3\theta }}{{{{(3\theta )}^2}}} \cdot {{(3\theta )}^2}}}{\text{ }}\]

\[ = \dfrac{4}{9} \cdot \dfrac{{\mathop {\lim }\limits_{\theta  \to 0} {{\left( {\dfrac{{\sin 2\theta }}{{2\theta }}} \right)}^2}}}{{\mathop {\lim }\limits_{\theta  \to 0} {{\left( {\dfrac{{\sin 3\theta }}{{3\theta }}} \right)}^2}}}\quad \left[ {\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1} \right]\]

$= \dfrac{4}{9}{\text{                            }}\left[ {x \to 0 \Rightarrow kx \to 0} \right]$

Correct Option:  A

Ans: Given:

\[\mathop {\lim }\limits_{\theta  \to 0} \dfrac{{\operatorname{co} {\text{sec }}x - \cot x}}{x}{\text{ }}\]

We know that,

$\because \mathop {\lim }\limits_{\theta  \to 0} \dfrac{{\tan \theta }}{\theta } = 1$

Let,

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\operatorname{cosec} x - \cot x}}{x}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{1}{{\sin x}} - \dfrac{{\cos x}}{{\sin x}}}}{x}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \cos x}}{{x \cdot \sin x}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{2{{\sin }^2}\dfrac{x}{2}}}{{x \cdot 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\tan \dfrac{x}{2}}}{x}$${\text{  }}\left[ {\because \mathop {\lim }\limits_{\theta  \to 0} \dfrac{{\tan \theta }}{\theta } = 1} \right]$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\tan \dfrac{x}{2}}}{{\dfrac{x}{2}}} \cdot \dfrac{1}{2}$

$= \dfrac{1}{2}$

Correct Option:  C

Ans: Given:

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{{\sqrt {x + 1}  - \sqrt {x - 1} }}$

We know that,

$\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$

Let,

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{{\sqrt {x + 1}  - \sqrt {1 - x} }}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{{\sqrt {x + 1}  - \sqrt {1 - x} }} \cdot \dfrac{{\sqrt {x + 1}  + \sqrt {1 - x} }}{{\sqrt {x + 1}  + \sqrt {1 - x} }}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x(\sqrt {x + 1}  + \sqrt {1 - x} )}}{{(x + 1) - (1 - x)}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x(\sqrt {x + 1}  + \sqrt {1 - x} )}}{{x + 1 - 1 + x}}$

$= \dfrac{1}{2}\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x}\mathop {\lim }\limits_{x \to 0} (\sqrt {x + 1}  + \sqrt {1 - x} )$

$= \dfrac{1}{2} \cdot 1 \cdot 2\,\left[ {\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1} \right]$

$= 1\quad$ 

Correct Option:  C

Ans: Given:

$\mathop {\lim }\limits_{x \to \pi /4} \dfrac{{{\text{se}}{{\text{c}}^2}{\text{ }}x - 2}}{{{\text{tan }}x - 1}}$

We know that,

\[\because \sec x = 1 + \tan x\]

\[\because {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]

Let,

$\mathop {\lim }\limits_{x \to \pi /4} \dfrac{{{{\sec }^2}x - 2}}{{\tan x - 1}}$

$= \mathop {\lim }\limits_{x \to \pi /4} \dfrac{{1 + {{\tan }^2}x - 2}}{{\tan x - 1}}$

$= \mathop {\lim }\limits_{x \to \pi /4} \dfrac{{{{\tan }^2}x - 1}}{{\tan x - 1}}$

$= \mathop {\lim }\limits_{x \to \pi /4} \dfrac{{(\tan x + 1)(\tan x - 1)}}{{(\tan x - 1)}}$

$= \mathop {\lim }\limits_{x \to \pi /4} (\tan x + 1)$

$= 2$

Correct Option:  D

Ans: Given:

$\mathop {\lim }\limits_{x \to 1} \dfrac{{(\sqrt x  - 1)(2x - 3)}}{{(2{x^2} + x - 3)}}$

We know that,

\[\because {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]

Let,

$\mathop {\lim }\limits_{x \to 1} \dfrac{{(\sqrt x  - 1)(2x - 3)}}{{2{x^2} + x - 3}}$

$= \mathop {\lim }\limits_{x \to 1} \dfrac{{(\sqrt x  - 1)(2x - 3)}}{{(2x + 3)(x - 1)}}$

$= \mathop {\lim }\limits_{x \to 1} \dfrac{{(\sqrt x  - 1)(2x - 3)}}{{(2x + 3)(\sqrt x  - 1)(\sqrt x  + 1)}}$

$= \mathop {\lim }\limits_{x \to 1} \dfrac{{2x - 3}}{{(2x + 3)(\sqrt x  + 1)}}$

$= \dfrac{{ - 1}}{{5 \times 2}}$

$= \dfrac{{ - 1}}{{10}}$

Correct Option:  B