JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Limits

The limit is a $\frac{0}{0}$ form at $x=\frac{\pi }{4}$, where $\cot x=\tan x=1$ and $\cos \left(x+\frac{\pi }{4}\right)=0$, so apply L'Hopital's rule. The numerator ${\cot }^{3}x-\tan x$ differentiates to $-3{\cot }^{2}x{\csc }^{2}x-{\sec }^{2}x$, which at $x=\frac{\pi }{4}$ is $-3(1)(2)-2=-8$. The denominator $\cos \left(x+\frac{\pi }{4}\right)$ differentiates to $-\sin \left(x+\frac{\pi }{4}\right)$, which at $x=\frac{\pi }{4}$ is $-\sin \frac{\pi }{2}=-1$. The limit is $\frac{-8}{-1}=8$.