Search any question & find its solution
Question:
Answered & Verified by Expert
If a function $f$ is defined by $f(x)=\frac{\cot ^3 x-\tan x}{\cos (x+\pi / 4)}(x \neq \pi / 4)$, then $\lim _{x \rightarrow \frac{\pi}{4}} f(x)=$
Options:
Solution:
1847 Upvotes
Verified Answer
The correct answer is:
8
$\lim _{x \rightarrow \frac{\pi}{4}} \frac{\cot ^3 x-\tan x}{\cos \left(x+\frac{\pi}{4}\right)}$
$\begin{aligned} & =\lim _{x \rightarrow \frac{\pi}{4}} \frac{\frac{1}{\tan 3}-\tan x}{\cos x \cos \frac{\pi}{4}-\sin x \sin \frac{\pi}{4}} \\ & =\lim _{x \rightarrow \frac{\pi}{4}} \frac{1-\tan ^4 x}{\tan ^3 x \cdot \frac{1}{\sqrt{2}}(\cos x-\sin x)} \\ & =\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sqrt{2}\left(1+\tan ^2 x\right)(1+\tan x)(1-\tan x)}{\tan ^2 x \cdot \tan x \cdot \cos x(1-\tan x)} \\ & =\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sqrt{2} \sec ^2 x(1+\tan x)}{\sin ^2 x \cdot \tan ^2 x} \\ & =2(\sqrt{2})^2(1+1)=8 .\end{aligned}$
$\begin{aligned} & =\lim _{x \rightarrow \frac{\pi}{4}} \frac{\frac{1}{\tan 3}-\tan x}{\cos x \cos \frac{\pi}{4}-\sin x \sin \frac{\pi}{4}} \\ & =\lim _{x \rightarrow \frac{\pi}{4}} \frac{1-\tan ^4 x}{\tan ^3 x \cdot \frac{1}{\sqrt{2}}(\cos x-\sin x)} \\ & =\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sqrt{2}\left(1+\tan ^2 x\right)(1+\tan x)(1-\tan x)}{\tan ^2 x \cdot \tan x \cdot \cos x(1-\tan x)} \\ & =\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sqrt{2} \sec ^2 x(1+\tan x)}{\sin ^2 x \cdot \tan ^2 x} \\ & =2(\sqrt{2})^2(1+1)=8 .\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.