Chapter 13: Limits and Derivatives

Ans: Given:

$\mathop {\lim }\limits_{x \to \pi /4} \dfrac{{\left( {\sin x - \cos x} \right)}}{{\left( {x - \dfrac{\pi }{4}} \right)}}{\text{   }}$

We know that,

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

$\because x \to \dfrac{\pi }{4} \Rightarrow \left( {x - \dfrac{\pi }{4}} \right) \to 0$

Then,

$\mathop {\lim }\limits_{x \to \pi /4} \dfrac{{\left( {\sin x - \cos x} \right)}}{{\left( {x - \dfrac{\pi }{4}} \right)}}{\text{ }}\left( {{\text{given}}} \right)$

$= \mathop {\lim }\limits_{x \to \pi /4} \dfrac{{\sqrt 2 \left( {\sin x \cdot \dfrac{1}{{\sqrt 2 }} - \cos x \cdot \dfrac{1}{{\sqrt 2 }}} \right)}}{{\left( {x - \dfrac{\pi }{4}} \right)}}$

$= \mathop {\lim }\limits_{x \to \pi /4} \dfrac{{\sqrt 2 \left( {\sin x\cos \dfrac{\pi }{4} - \cos x \cdot \sin \dfrac{\pi }{4}} \right)}}{{\left( {x - \dfrac{\pi }{4}} \right)}}$

$= \mathop {\lim }\limits_{x \to \pi /4} \dfrac{{\sqrt 2 \left\{ {\sin \left( {x - \dfrac{\pi }{4}} \right)} \right\}}}{{\left( {x - \dfrac{\pi }{4}} \right)}}$

$= \sqrt 2 \mathop {\lim }\limits_{x \to \pi /4} \dfrac{{\sin \left( {x - \dfrac{\pi }{4}} \right)}}{{\left( {x - \dfrac{\pi }{4}} \right)}}{\text{  }}\left[ {\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1} \right]$

$= \sqrt 2 \quad {\text{                          }}\left[ {\because x \to \dfrac{\pi }{4} \Rightarrow \left( {x - \dfrac{\pi }{4}} \right) \to 0} \right]$

Ans: Given:

$\mathop {\lim }\limits_{x \to \pi /6} \dfrac{{\sqrt 3 \sin x - \cos x}}{{x - \dfrac{\pi }{6}}}$

We know that,

$\because \sin A\cos B - \cos A\sin B = \sin (A - B)]$

$\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1{\text{ and }}x \to \dfrac{\pi }{6} \Rightarrow \left( {x - \dfrac{\pi }{6}} \right) \to 0$

Then,

$= \mathop {\lim }\limits_{x \to \pi /6} \dfrac{{\sqrt 3 \sin x - \cos x}}{{x - \dfrac{\pi }{6}}}{\text{ }}\left( {{\text{given}}} \right)$

$= \mathop {\lim }\limits_{x \to \pi /6} \dfrac{{2\left( {\dfrac{{\sqrt 3 }}{2}\sin x - \dfrac{1}{2}\cos x} \right)}}{{\left( {x - \dfrac{\pi }{6}} \right)}}$

$= \mathop {\lim }\limits_{x \to \pi /6} \dfrac{{2\left( {\sin x\cos \dfrac{\pi }{6} - \cos x\sin \dfrac{\pi }{6}} \right)}}{{\left( {x - \dfrac{\pi }{6}} \right)}}{\text{  }}[\because \sin A\cos B - \cos A\sin B = \sin (A - B)]$

$= 2\mathop {\lim }\limits_{x \to \pi /6} \dfrac{{\sin \left( {x - \dfrac{\pi }{6}} \right)}}{{\left( {x - \dfrac{\pi }{6}} \right)}}$

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

$\left[ {\because x \to \dfrac{\pi }{6} \Rightarrow \left( {x - \dfrac{\pi }{6}} \right) \to 0} \right]$ 

$= 2$

Ans: Given:

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 2x + 3x}}{{2x + \tan 3x}}$

We know that,

\[\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1\,{\text{and }}\mathop {\lim }\limits_{x \to 0} \dfrac{{\tan x}}{x} = 1\]

Then,

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 2x + 3x}}{{2x + \tan 3x}}{\text{ }}\left( {{\text{given}}} \right)$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{{\sin 2x + 3x}}{{2x}} \cdot 2x}}{{\dfrac{{2x + \tan 3x}}{{3x}} \cdot 3x}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\dfrac{{\sin 2x}}{{2x}} + \dfrac{{3x}}{{2x}}} \right)2x}}{{\left( {\dfrac{{2x}}{{3x}} + \dfrac{{\tan 3x}}{{3x}}} \right)3x}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{{\sin 2x}}{{2x}} + \dfrac{3}{2}}}{{\dfrac{2}{3} + \dfrac{{\tan 3x}}{{3x}}}} \cdot \dfrac{2}{3}$

$= \dfrac{2}{3}\mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{{\sin 2x}}{{2x}} + \dfrac{3}{2}}}{{\dfrac{2}{3} + \mathop {\lim }\limits_{x \to 0} \dfrac{{\tan 3x}}{{3x}}}}$

\[ = \dfrac{2}{3}\left( {\dfrac{{1 + \dfrac{3}{2}}}{{\dfrac{2}{3} + 1}}} \right)\quad \left[ {\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1} \right.\,{\text{and }}\left. {\mathop {\lim }\limits_{x \to 0} \dfrac{{\tan x}}{x} = 1} \right]\]

$= \dfrac{2}{3} \times \dfrac{{\dfrac{5}{5}}}{{\dfrac{5}{3}}} = \dfrac{2}{3} \times \dfrac{5}{2} \times \dfrac{3}{5}$

$= 1$ 

Ans: Given:

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

We know that,

$\because \sin C - \sin D = 2\cos \dfrac{{C + D}}{2} \cdot \sin \dfrac{{C - D}}{2}$

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

Then,

\[ = \mathop {\lim }\limits_{x \to a} \dfrac{{\sin x - \sin a}}{{\sqrt x  - \sqrt a }}{\text{ }}\left( {{\text{given}}} \right)\]

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{2\cos \left( {\dfrac{{x + a}}{2}} \right)\sin \left( {\dfrac{{x - a}}{2}} \right)}}{{\sqrt x  - \sqrt a }}{\text{ }}\left[ {\because \sin C - \sin D = 2\cos \dfrac{{C + D}}{2} \cdot \sin \dfrac{{C - D}}{2}} \right]$

$= \mathop {\lim }\limits_{x \to a} \dfrac{{2\cos \left( {\dfrac{{x + a}}{2}} \right)\sin \left( {\dfrac{{x - a}}{2}} \right)(\sqrt {x + \sqrt a } )}}{{(\sqrt {x - \sqrt a )} (\sqrt x  + \sqrt a )}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{2\cos \left( {\dfrac{{x + a}}{2}} \right)\sin \left( {\dfrac{{x - a}}{2}} \right)(\sqrt x  + \sqrt a )}}{{x - a}}$

$= 2\mathop {\lim }\limits_{x \to a} \cos \left( {\dfrac{{x + a}}{2}} \right)(\sqrt x  + \sqrt a )\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \left( {\dfrac{{x - a}}{2}} \right)}}{{2\left( {\dfrac{{x - a}}{2}} \right)}}$

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

$= 2 \cdot \cos \dfrac{a}{2} \cdot \sqrt a  \cdot \dfrac{1}{2}\quad {\text{                 }}\left[ {\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1} \right]$

$= \sqrt a \cos \dfrac{a}{2}$

Ans: Given:

$\mathop {\lim }\limits_{x \to \pi /6} \dfrac{{{{\cot }^2}x - 3}}{{\operatorname{cosec} x - 2}}$

We know that,

$\because {\operatorname{cosec} ^2}x = 1 + {\cot ^2}x$

Then,

$= \mathop {\lim }\limits_{x \to \pi /6} \dfrac{{{{\cot }^2}x - 3}}{{\operatorname{cosec} x - 2}}{\text{ }}\left( {{\text{given}}} \right)$

$= \mathop {\lim }\limits_{x \to \pi /6} \dfrac{{{{\operatorname{cosec} }^2}x - 1 - 3}}{{\operatorname{cosec} x - 2}}\quad \left[ {\because {{\operatorname{cosec} }^2}x = 1 + {{\cot }^2}x} \right]$

$= \mathop {\lim }\limits_{x \to \pi /6} \dfrac{{{{\operatorname{cosec} }^2}x - 4}}{{\operatorname{cosec} x - 2}}$

$= \mathop {\lim }\limits_{x \to \pi /6} \dfrac{{{{(\operatorname{cosec} x)}^2} - {{(2)}^2}}}{{\operatorname{cosec} x - 2}}$

$= \mathop {\lim }\limits_{x \to \pi /6} \dfrac{{(\operatorname{cosec} x + 2)(\operatorname{cosec} x - 2)}}{{(\operatorname{cosec} x - 2)}}$

$= \mathop {\lim }\limits_{x \to \pi /6} (\operatorname{cosec} x + 2)$

$= \operatorname{cosec} \dfrac{\pi }{6} + 2 = 2 + 2$

$= 4$

Ans: Given:

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

We know that,

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

$\because \sin x = 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$

Then,

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

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

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt 2 \left( {1 - \cos \dfrac{x}{2}} \right)}}{{{{\sin }^2}x}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt 2 \left( {1 - 1 + 2{{\sin }^2}\dfrac{x}{4}} \right)}}{{{{\sin }^2}x}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt 2 \left( {2{{\sin }^2}\dfrac{x}{4}} \right)}}{{{{\sin }^2}x}}$

$= \mathop {\lim }\limits_{x \to 0} 2\sqrt 2 \dfrac{{{{\sin }^2}\dfrac{x}{4}}}{{{{\left( {\dfrac{x}{4}} \right)}^2}}} \cdot \dfrac{{{{\left( {\dfrac{x}{4}} \right)}^2}}}{{{{\sin }^2}x}}$

$= 2\sqrt 2 \mathop {\lim }\limits_{x \to 0} {\left( {\dfrac{{\sin \dfrac{x}{4}}}{{\dfrac{x}{4}}}} \right)^2} \cdot \mathop {\lim }\limits_{x \to 0} {\left( {\dfrac{x}{{\sin x}}} \right)^2} \cdot \dfrac{1}{{16}}$

$= 2\sqrt 2  \cdot 1 \cdot 1 \cdot \dfrac{1}{{16}}$

$= \dfrac{1}{{4\sqrt 2 }}$

Ans: Given:

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x - 2\sin 3x + \sin 5x}}{x}$

We know that,

$\because \sin x + \sin y = 2\cos \dfrac{{x + y}}{2}\cos \dfrac{{x - y}}{2}$

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

Then,

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x - 2\sin 3x + \sin 5x}}{x}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 5x + \sin x - 2\sin 3x}}{x}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{2\sin 3x\cos 2x - 2\sin 3x}}{x}$$\left[ {\because \sin x + \sin y = 2\cos \dfrac{{x + y}}{2}\cos \dfrac{{x - y}}{2}} \right]$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{2\sin 3x(\cos 2x - 1)}}{x}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{2\sin 3x}}{{\dfrac{1}{3} \times 3x}}(\cos 2x - 1)$

$= 6\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 3x}}{{3x}}(\cos 2x - 1)$

$= 6\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 3x}}{{3x}} \cdot \mathop {\lim }\limits_{x \to 0} (\cos 2x - 1)$$\left[ {\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1} \right]$

$= 6 \times 1 \times 0 = 0$

Ans: Given:

$\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^4} - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to k} \dfrac{{{x^3} - {k^3}}}{{{x^2} - {k^2}}}$

We know that,

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

$\because \mathop {\lim }\limits_{x \to a} \dfrac{{f(x)}}{{g(x)}} = \dfrac{{\mathop {\lim }\limits_{x \to a} f(x)}}{{\mathop {\lim }\limits_{x \to a} g(x)}}$

Then,

$\Rightarrow \quad \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^4} - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to k} \dfrac{{{x^3} - {k^3}}}{{{x^2} - {k^2}}}\,\left( {{\text{given}}} \right)$

$\Rightarrow \quad 4{(1)^{4 - 1}} = \mathop {\lim }\limits_{x \to k} \dfrac{{\dfrac{{{x^3} - {k^3}}}{{x - k}}}}{{\dfrac{{{x^2} - {k^2}}}{{x - k}}}}\quad \left[ {\because \mathop {\lim }\limits_{x \to a} \dfrac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}} \right]$

$\Rightarrow \quad 4 = \dfrac{{\mathop {\lim }\limits_{x \to k} \dfrac{{{x^3} - {k^3}}}{{x - k}}}}{{\mathop {\lim }\limits_{x \to k} \dfrac{{{x^2} - {k^2}}}{{x - k}}}}1\quad \left[ {\because \mathop {\lim }\limits_{x \to a} \dfrac{{f(x)}}{{g(x)}} = \dfrac{{\mathop {\lim }\limits_{x \to a} f(x)}}{{\mathop {\lim }\limits_{x \to a} g(x)}}} \right]$

$\Rightarrow \quad 4 = \dfrac{{3{k^2}}}{{2k}}$

$\Rightarrow {\text{ }}4 = \dfrac{3}{2}k$

$\therefore \quad k = \dfrac{{4 \times 2}}{3} = \dfrac{8}{3}$

Differentiate each of the functions w. r. t x  in Exercises 29 to 42:

Ans: Given:

\[\dfrac{{{x^4} + {x^3} + {x^2} + 1}}{x}\]

Then

\[ = \dfrac{d}{{dx}}\left( {\dfrac{{{x^4} + {x^3} + {x^2} + 1}}{x}} \right)\]

\[ = \dfrac{d}{{dx}}\left( {{x^3} + {x^2} + x + \dfrac{1}{x}} \right)\]

\[ = \dfrac{d}{{dx}}{x^3} + \dfrac{d}{{dx}}{x^2} + \dfrac{d}{{dx}}x + \dfrac{d}{{dx}}\left( {\dfrac{1}{x}} \right)\]          (by chain rule)

\[ = 3{x^2} + 2x + 1 + \left( { - \dfrac{1}{{{x^2}}}} \right)\]

\[ = 3{x^2} + 2x + 1 - \dfrac{1}{{{x^2}}}\]

\[ = \dfrac{{3{x^4} + 2{x^3} + {x^2} - 1}}{{{x^2}}}\]

Ans: Given:

${\left( {x + \dfrac{1}{x}} \right)^3}$

${\text{Let }}\quad y = {\left( {x + \dfrac{1}{x}} \right)^3}$

Differentiate on both side w.r.t \[x\]

$\therefore \quad \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}{\left( {x + \dfrac{1}{x}} \right)^3}$

$= 3{\left( {x + \dfrac{1}{x}} \right)^{3 - 1}}\dfrac{d}{{dx}}\left( {x + \dfrac{1}{x}} \right)$ (by chain rule)

$= 3{\left( {x + \dfrac{1}{x}} \right)^2}\left( {1 - \dfrac{1}{{{x^2}}}} \right)$

$= 3{x^2} - \dfrac{3}{{{x^2}}} - \dfrac{3}{{{x^4}}} + 3$