Chapter 13: Limits and Derivatives

Ans: Given:

\[\cos \left( {{x^2} + 1} \right)\]

We know that, To differentiate the given equation

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

$\because x \to 0 \Rightarrow kx \to 0$

Let,

\[\therefore \dfrac{d}{{dx}}f\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\]

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

$= \mathop {\lim }\limits_{h \to 0} \dfrac{{ - 2\sin \left\{ {\dfrac{{{{(x + h)}^2} + 1 + {x^2} + 1}}{2}} \right\}\sin \left\{ {\dfrac{{{{(x + h)}^2} + 1 - {x^2} - 1}}{2}} \right\}}}{h}\left[ {\because \cos C - \cos D =  - 2\sin \dfrac{{C + D}}{2} \cdot \sin \dfrac{{C - D}}{2}} \right]$$= \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h}\left[ { - 2\sin \left\{ {\dfrac{{{{(x + h)}^2} + {x^2} + 2}}{2}} \right\}\sin \left\{ {\dfrac{{{{(x + h)}^2} - {x^2}}}{2}} \right\}} \right]$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h}\left[ { - 2\sin \left\{ {\dfrac{{{{(x + h)}^2} + {x^2} + 2}}{2}} \right\}\sin \left\{ {\dfrac{{{x^2} + {h^2} + 2xh - {x^2}}}{2}} \right\}} \right]$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h}\left[ { - 2\sin \left\{ {\dfrac{{{{(x + h)}^2} + {x^2} + 2}}{2}} \right\}\sin \left\{ {\dfrac{{{h^2} + 2hx}}{2}} \right\}} \right]$

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

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

\[ =  - 2x\sin \left( {{x^2} + 1} \right)\quad \left[ {\because x \to 0 \Rightarrow kx \to 0} \right]\]

Ans: Given, $\dfrac{{ax + b}}{{cx + d}}$

Let,

 $f\left( x \right) = \dfrac{{ax + b}}{{cx + d}}$

$f\left( {x + h} \right) = \dfrac{{a\left( {x + h} \right) + b}}{{c\left( {x + h} \right) + d}}$

$\therefore \dfrac{d}{{dx}}f\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h}\left[ {f\left( {x + h} \right) - f\left( x \right)} \right]$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h}\left[ {\dfrac{{a(x + h) + b}}{{c(x + h) + d}} - \dfrac{{ax + b}}{{cx + d}}} \right]$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h}\left[ {\dfrac{{ax + b + ah}}{{c(x + h) + d}} - \dfrac{{ax + b}}{{cx + d}}} \right]$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h}\left[ {\dfrac{{(ax + ah + b)(cx + d) - (ax + b\{ c(x + h) + d\} }}{{\{ c(x + h) + d\} (cx + d)}}} \right]$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h}\left[ {\dfrac{{(ax + ah + b)(cx + d) - (ax + b)(cx + ch + d)}}{{\{ c(x + h) + d)\} (cx + d)}}} \right]$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h}\left[ {\dfrac{{ac{x^2} + achx + bcx + adx + adh + bd - ac{x^2} - achx - adx - bcx - bch - bd}}{{\{ c(x + h) + d\} (cx + d)}}} \right]$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h}\left[ {\dfrac{{adh - bch}}{{\{ c(x + h) + d\} (cx + d)}}} \right]$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{{ad - bc}}{{\{ c(x + h) + d\} (cx + d)}}$

$= \dfrac{{ad - bc}}{{{{(cx + d)}^2}}}$

Ans: Given:

${x^{\dfrac{2}{3}}}$

We know that,

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

Let,

\[f(x) = {x^{2/3}}\]

\[f(x + h) = {(x + h)^{2/3}}\]

\[\dfrac{d}{{dx}}f(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}\]

\[ = \mathop {\lim }\limits_{(h \to 0} \left[ {\dfrac{{(x + h)2/3 - {x^{2/3}}}}{h}} \right]\]

\[ = \mathop {\lim }\limits_{(x + h) \to x} \left[ {\dfrac{{(x + h)2/3 - {x^{2/3}}}}{{(x + h) - x}}} \right]\left[ {\because \mathop {\lim }\limits_{x \to a} \dfrac{{{x^x} - {a^n}}}{{x - a}} = n{a^{n - 1}}} \right]\]

\[ = \dfrac{2}{3}{(x)^{2/3 - 1}}\quad \]

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

Ans: Given:

\[x\cos x\]

We know that,

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

Then,

$= \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h}[(x + h)\cos (x + h) - x\cos x]$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h}[x\cos (x + h) + h\cos (x + h) - x\cos x]$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h}[x\{ \cos (x + h) - \cos x\}  + h\cos (x + h)]$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h}\left[ {x\left\{ { - 2\sin \left( {\dfrac{{2x + h}}{2}} \right)\sin \dfrac{h}{2}} \right\} + h\cos (x + h)} \right]$

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

\[ =  - 2\mathop {\lim }\limits_{h \to 0} x\sin \left( {x + \dfrac{h}{2}} \right)\mathop {\lim }\limits_{h \to 0} \dfrac{{\sin \dfrac{h}{2}}}{{\dfrac{h}{2}}} \cdot \dfrac{1}{2} + \mathop {\lim }\limits_{h \to 0} \cos (x + h)\]

\[ =  - 2 \cdot \dfrac{1}{2}x\sin x + \cos x\]

\[ = \cos x - x\sin x\]

Evaluate each of the following limits in Exercises 47 to 53:

Ans: Given:

$\mathop {\lim }\limits_{y \to 0} \dfrac{{\left( {x + y} \right){\text{sec}}\left( {x + y} \right) - x\sec x}}{y}$

Let,

$\mathop {\lim }\limits_{y \to 0} \dfrac{{\left( {x + y} \right){\text{sec}}\left( {x + y} \right) - x\sec x}}{y}$

$= \mathop {\lim }\limits_{y \to 0} \dfrac{{\dfrac{{x + y}}{{\cos (x + y)}} - \dfrac{x}{{\cos x}}}}{y}$

$= \mathop {\lim }\limits_{y \to 0} \dfrac{{(x + y)\cos x - x\cos (x + y)}}{{y\cos x\cos (x + y)}}$

$= \mathop {\lim }\limits_{y \to 0} \left[ {\dfrac{{x\cos x + y\cos x - x\cos (x + y)}}{{y\cos x\cos (x + y)}}} \right]$

$= \mathop {\lim }\limits_{y \to 0} \left[ {\dfrac{{x\cos x - x\cos (x + y) + y\cos x}}{{y\cos x\cos (x + y)}}} \right]$

$= \mathop {\lim }\limits_{y \to 0} \dfrac{{x\{ \cos x - \cos (x + y)\}  + y\cos x}}{{y\cos x\cos (x + y)}}$

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

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

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

$\quad \left[ {\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1} \right.$ and $\left. {x \to 0 \Rightarrow kx \to 0} \right]$

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

$= \dfrac{{2x\sin x}}{{\cos x\cos x}} \cdot \dfrac{1}{2} + \sec x$

$= x\tan x\sec x + \sec x$

$= \sec x(x\tan x + 1)$

Ans: Given:

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\left[ {\sin \left( {\alpha  + \beta } \right)x + \sin \left( {\alpha  - \beta } \right)x + \sin 2\alpha x} \right]}}{{\cos 2\beta x - \cos 2\alpha x}} \cdot x$

We know that,

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

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

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

Let,

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\left[ {\sin \left( {\alpha  + \beta } \right)x + \sin \left( {\alpha  - \beta } \right)x + \sin 2\alpha x} \right]}}{{\cos 2\beta x - \cos 2\alpha x}} \cdot x$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\left[ {\sin \left( {\alpha  + \beta } \right)x + \sin \left( {\alpha  - \beta } \right)x + \sin 2\alpha x} \right]}}{{\cos 2\beta x - \cos 2\alpha x}} \cdot x$

$\left[ {\because \sin C + \sin D = 2\sin \dfrac{{C + D}}{2}\cos \dfrac{{C - D}}{2}} \right]$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\left[ {2\sin \alpha x \cdot \cos \beta x + \sin 2\alpha x} \right]}}{{\cos 2\beta x - \cos 2\alpha x}} \cdot x$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{[2\sin \alpha x\cos \beta x + \sin 2\alpha x]x}}{{2\sin (\alpha  + \beta )x\sin (\alpha  - \beta )x}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{[2\sin \alpha x\cos \beta x + 2\sin \alpha x\cos \alpha x]x}}{{2\sin (\alpha  + \beta )x\sin (\alpha  - \beta )x}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{2\sin \alpha x[\cos \beta x + \cos \alpha x]x}}{{2\sin (\alpha  + \beta )x\sin (\alpha  - \beta )x}}$

$\left[ {\because \cos C - \cos D = 2\sin \dfrac{{C + D}}{2} \cdot \sin \dfrac{{D - C}}{2}} \right]$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \alpha x\left[ {2\cos \left( {\dfrac{{\alpha  + \beta }}{2}} \right)x\cos \left( {\dfrac{{\alpha  - \beta }}{2}} \right)x} \right]x}}{{2\sin \left( {\dfrac{{\alpha  + \beta }}{2}} \right)x\cos \left( {\dfrac{{\alpha  + \beta }}{2}} \right)x \cdot 2\sin \left( {\dfrac{{\alpha  - \beta }}{2}} \right)x\cos \left( {\dfrac{{\alpha  - \beta }}{2}} \right)x}}$

$\left[ {\because \cos C + \cos D = 2\cos \dfrac{{C + D}}{2}\cos \dfrac{{C - D}}{2}\operatorname{and} \sin 2\theta  = 2\sin \theta \cos \theta } \right]$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \alpha x \cdot x}}{{2\sin \left( {\dfrac{{\alpha  + \beta }}{2}} \right)x\sin \left( {\dfrac{{\alpha  - \beta }}{2}} \right)x}}$

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

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

$= \dfrac{1}{2} \cdot \dfrac{{\alpha  \cdot 4}}{{{\alpha ^2} - {\beta ^2}}}\left[ {\dfrac{{\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \alpha x}}{{\alpha x}}}}{{\mathop {\lim }\limits_{x \to 0} \sin \dfrac{{\left( {\dfrac{{\alpha  + \beta }}{2}} \right)x}}{{\left( {\dfrac{{\alpha  + \beta }}{2}} \right)x}}\mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\dfrac{{\alpha  - \beta }}{2}} \right)x}}{{\left( {\dfrac{{\alpha  - \beta }}{2}} \right)x}}}}} \right]$

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

$= \dfrac{1}{2} \cdot \dfrac{{4\alpha }}{{{\alpha ^2} - {\beta ^2}}}$

$= \dfrac{{2\alpha }}{{{\alpha ^2} - {\beta ^2}}}$

Ans: Given:

$\mathop {\lim }\limits_{x \to \pi /4} \dfrac{{{{\tan }^3}x - \tan x}}{{\cos \left( {x + \dfrac{\pi }{4}} \right)}}$

We know that,

$\dfrac{0}{0}{\text{ form}}$

$\because {a^2} - {b^2} = (a + b)(a - b)$

$\because \cos A \cdot \cos B - \sin A\sin B = \cos (A + B)$

Let,

$\mathop {\lim }\limits_{x \to \pi /4} \dfrac{{{{\tan }^3}x - \tan x}}{{\cos \left( {x + \dfrac{\pi }{4}} \right)}}$$\left[ {\dfrac{0}{0}{\text{ form}}} \right]$

$= \mathop {\lim }\limits_{x \to \pi /4} \dfrac{{\tan x\left( {{{\tan }^2}x - 1} \right)}}{{\cos \left( {x + \dfrac{\pi }{4}} \right)}} = \mathop {\lim }\limits_{x \to \pi /4} \tan x \cdot \mathop {\lim }\limits_{x \to \pi /4} \left( {\dfrac{{1 - {{\tan }^2}x}}{{\cos \left( {x + \dfrac{{g\pi }}{4}} \right)}}} \right)$

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

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

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

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

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

$\because \cos A \cdot \cos B - \sin A\sin B = \cos (A + B)$

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

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

$=  - 2\sqrt 2  \times \sqrt 2$

$=  - 4$

Ans: Given:

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

We know that,

$\because {\sin ^2}\theta  + {\cos ^2}\theta  = \sin 2\theta  = 2\sin \theta \cos \theta$

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

$\because {a^2} - {b^2} = (a + b)(a - b]$

Let,

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

$= \mathop {\lim }\limits_{x \to \pi } \dfrac{{{{\cos }^2}\dfrac{x}{4} + {{\sin }^2}\dfrac{x}{4} - 2 \cdot \sin \dfrac{x}{4} \cdot \cos \dfrac{x}{4}}}{{\cos \dfrac{x}{2} \cdot \left( {\cos \dfrac{x}{4} - \sin \dfrac{x}{4}} \right)}}\left[ {\because {{\sin }^2}\theta  + {{\cos }^2}\theta  = 1\sin 2\theta  = 2\sin \theta \cos \theta } \right]$

$= \mathop {\lim }\limits_{x \to \pi } \dfrac{{{{\left( {\cos \dfrac{x}{4} - \sin \dfrac{x}{4}} \right)}^2}}}{{\left( {{{\cos }^2}\dfrac{x}{4} - {{\sin }^2}\dfrac{x}{4}} \right)\left( {\cos \dfrac{x}{4} - \sin \dfrac{x}{4}} \right)}}\quad \left[ {\because {{\cos }^2}2\theta  = {{\cos }^2}\theta  - {{\sin }^2}\theta } \right]$

$= \mathop {\lim }\limits_{x \to \pi } \dfrac{{\left( {\cos \dfrac{x}{4} - \sin \dfrac{x}{4}} \right)}}{{\left( {\cos \dfrac{x}{4} + \sin \dfrac{x}{4}} \right)\left( {\cos \dfrac{x}{4} - \sin \dfrac{x}{4}} \right)}}\quad \left[ {\because {a^2} - {b^2} = (a + b)(a - b} \right]$

$= \mathop {\lim }\limits_{x \to \pi } \dfrac{1}{{\cos \dfrac{x}{4} + \sin \dfrac{x}{4}}}$

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

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

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

Ans: Given:

$\mathop {\lim }\limits_{x \to \pi /4} \dfrac{{|x - 4|}}{{x - 4}}$

Let,

$\mathop {\lim }\limits_{x \to \pi /4} \dfrac{{|x - 4|}}{{x - 4}}$

${\text{LHL}} = \mathop {\lim }\limits_{x \to \dfrac{{{\pi ^ - }}}{4}} \dfrac{{ - (x - 4)}}{{x - 4}}\quad [\because |x - 4| =  - (x - 4),x < 4]$

${\text{       }} =  - 1$

${\text{RHL}} = \mathop {\lim }\limits_{x \to \dfrac{{{\pi ^ + }}}{4}} \dfrac{{(x - 4)}}{{x - 4}}[\because |x - 4| = (x - 4),x > 4]$

${\text{        }} = 1$

$\therefore {\text{LHL}} \ne {\text{RHL}}$

So, limit does not exist

Ans:

$\therefore {\text{LHL}} = \mathop {\lim }\limits_{x \to \dfrac{{{\pi ^ - }}}{2}} \dfrac{{k\cos x}}{{\pi  - 2x}}$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{{k\cos \left( {\dfrac{\pi }{2} - h} \right)}}{{\pi  - 2\left( {\dfrac{\pi }{2} - h} \right)}}$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{{k\sin h}}{{\pi  - \pi  + 2h}}$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{{{\text{ksin}}h}}{{2h}}$

$= \dfrac{k}{2}\mathop {\lim }\limits_{h \to 0} \dfrac{{\sin h}}{h}$

$= \dfrac{k}{2} \cdot 1$

$= \dfrac{k}{2}$

${\text{RHL}} = \mathop {\lim }\limits_{x \to \dfrac{{{\pi ^ + }}}{2}} \dfrac{{k\cos x}}{{\pi  - 2x}}$

$= \mathop {\lim }\limits_{x \to \dfrac{\pi }{2}}  + \dfrac{{k\cos \left( {\dfrac{\pi }{2} + h} \right)}}{{\pi  - 2\left( {\dfrac{\pi }{2} + h} \right)}}$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{{ - k\sin h}}{{\pi  - \pi  - 2h}}$

$= \mathop {\lim }\limits_{h \to 0} \dfrac{{k\sin h}}{{2h}}$

$= \dfrac{k}{2}\mathop {\lim }\limits_{h \to 0} \dfrac{{\sin h}}{h}$

$= \dfrac{k}{2}$

${\text{ }}f\left( {\dfrac{\pi }{2}} \right) = 3$ (given)

It is given that,

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

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

$\therefore k = 6$