Question

Evaluate the limit:
$\underset{x\to \frac{\pi }{8}}{\text{lim}}\frac{\cot 4x-\cos 4x}{(\pi -8x{)}^3}$

Answer 1

We have,

$\underset{x\to \frac{\pi }{8}}{\text{lim}}\frac{\cot 4x-\cos 4x}{(\pi -8x{)}^3}$

Put $x=\frac{\pi }{8}+h,h\to 0$

$=\underset{h\to 0}{\text{lim}}\frac{\cot 4(\frac{\pi }{8}+h)-\cos 4(\frac{\pi }{8}+h)}{{(\pi -8(\frac{\pi }{8}+h))}^3}$

$=\underset{h\to 0}{\text{lim}}\frac{\cot (\frac{\pi }{2}+4h)-\cos (\frac{\pi }{2}+4h)}{(\pi -\pi -8h{)}^3}$

$=\underset{h\to 0}{\text{lim}}\frac{-\tan 4h+\sin 4h}{(-8h{)}^3}$

$[\because \cot (\frac{\pi }{2}+\theta )=-\tan \theta ]$

$=\underset{h\to 0}{\text{lim}}\frac{-\frac{\sin 4h}{\cos 4h}+\sin 4h}{(-8h{)}^3}$

$[\because \cos (\frac{\pi }{2}+\theta )=-\sin \theta ]$

$=\underset{h\to 0}{\text{lim}}\frac{\sin 4h(1-\frac{1}{\cos 4h})}{(-8h{)}^3}$

$=\underset{h\to 0}{\text{lim}}\frac{\sin 4h\left(\frac{\cos 4h-1}{\cos 4h}\right)}{(-8h{)}^3}$

$=\underset{h\to 0}{\text{lim}}\frac{\sin 4h\left(\frac{-(1-\cos 2(2h\left)\right)}{\cos 4h}\right)}{(-8h{)}^3}$

$=\underset{h\to 0}{\text{lim}}\frac{\sin 4h\left(\frac{-(1-\cos 2(2h\left)\right)}{\cos 4h}\right)}{(-8h{)}^3}$

$=\underset{h\to 0}{\text{lim}}\frac{\sin 4h\left(\frac{-2{\sin }^{2}2h}{\cos 4h}\right)}{-8\times 8\times 8h^3}$

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

$=\underset{h\to 0}{\text{lim}}\frac{{\sin }^{2}2h\left(\frac{\sin 4h}{\cos 4h}\right)}{4\times 8\times 8h^3}$

$=\underset{h\to 0}{\text{lim}}\frac{{\sin }^{2}2h\left(\tan 4h\right)}{4\times 8\times 8h^3}$

$=\frac{1}{16}\underset{h\to 0}{\text{lim}}\frac{{\sin }^{2}2h}{(2h{)}^2}\times \frac{\tan 4h}{4h}$

$=\frac{1}{16}\times 1\times 1$

$[\because \underset{x\to 0}{\text{lim}}\frac{\sin x}{x}=1,\underset{x\to 0}{\text{lim}}\frac{\tan x}{x}=1]$

$=\frac{1}{16}$

Therefore,

$\underset{x\to \frac{\pi }{8}}{\text{lim}}\frac{\cot 4x-\cos 4x}{(\pi -8x{)}^3}=\frac{1}{16}$