JEE Main 2019 — Limits Question with Solution

$=\underset{x\to \frac{\pi }{4}}{\lim }\frac{\left({\displaystyle \frac{1}{{\tan }^{3}x}}-\tan x\right)}{\left(\frac{1}{\sqrt{2}}\cos x-\frac{1}{\sqrt{2}}\sin x\right)} \left[\because \cot \theta =\frac{1}{\tan \theta }\right]$

$=\underset{x\to \frac{\pi }{4}}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{\left(1-\mathrm{t}\mathrm{a}{\mathrm{n}}^{4}x\right)}{\mathrm{t}\mathrm{a}{\mathrm{n}}^{3}x.\left(\frac{1}{\sqrt{2}}\mathrm{c}\mathrm{o}\mathrm{s}x-\frac{1}{\sqrt{2}}\mathrm{s}\mathrm{i}\mathrm{n}x\right)}$

$=\underset{x\to \frac{\pi }{4}}{\lim }\frac{\left(1-{\tan }^{2}x\right)\left(1+{\tan }^{2}x\right)}{{\tan }^{3}x.\left(\frac{1}{\sqrt{2}}\cos x-\frac{1}{\sqrt{2}}\sin x\right)}$

$=\underset{x\to \frac{\pi }{4}}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{\sqrt{2}\left[\left(1-\tan x\right)\left(1+\tan x\right){\sec }^{2}x\right]}{\mathrm{t}\mathrm{a}{\mathrm{n}}^{3}x.\left(\mathrm{c}\mathrm{o}\mathrm{s}x-\mathrm{s}\mathrm{i}\mathrm{n}x\right)} \left[\because 1+{\tan }^{2}x={\sec }^{2}x \& a^2-b^2=\left(a-b\right)\left(a+b\right)\right]$

$=\underset{x\to \frac{\pi }{4}}{\lim }\frac{\sqrt{2}\left[\left(1-{\displaystyle \frac{\sin x}{\cos x}}\right)\left(1+\tan x\right){\sec }^{2}x\right]}{{\tan }^{3}x.\left(\cos x-\sin x\right)}$

$=\underset{x\to \frac{\pi }{4}}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{\sqrt{2}\left(\cos x-\sin x\right).\left(1+\tan x\right).\mathrm{s}\mathrm{e}{\mathrm{c}}^{2}x}{\mathrm{c}\mathrm{o}\mathrm{s}x.\mathrm{t}\mathrm{a}{\mathrm{n}}^{3}x.\left(\cos x-\sin x\right)}$

$=\underset{x\to \frac{\pi }{4}}{\lim }\frac{\sqrt{2}\left(1+\tan x\right)}{\cos x.{\displaystyle \frac{{\sin }^{3}x}{{\cos }^{3}x}}.{\cos }^{2}x}$

$=\underset{x\to \frac{\pi }{4}}{\lim }\frac{\sqrt{2}\left(1+\tan x\right)}{{\sin }^{3}x}$

$=\frac{\sqrt{2}\left(1+\tan {\displaystyle \frac{\pi }{4}}\right)}{{\sin }^3{\displaystyle \frac{\pi }{4}}}$

$=8$.