​$$\text { Find } \lim _{x \rightarrow \frac{\pi}{4}} \frac{\cot ^3 x-\tan x}{\cos \left(x+\frac{\pi}{4}\right)}.$$​​

Correct option is B

​$$\begin{aligned}&\text{At } x = \frac{\pi}{4}, \text{ we have:} \\&\quad \cot x = \tan^{-1} x = 1 \Rightarrow \cot^3 x = 1 \\&\quad \tan x = 1 \\&\quad \cos\left(x + \frac{\pi}{4} \right) = \cos\left(\frac{\pi}{2} \right) = 0 \\[10pt]&\text{So, the limit becomes } \frac{0}{0}, \text{ which is indeterminate. Apply L'Hôpital's Rule:} \\&\text{Differentiate numerator:} \\&\frac{d}{dx}(\cot^3 x - \tan x) = \frac{d}{dx}(\cot^3 x) - \frac{d}{dx}(\tan x) \\&= 3 \cot^2 x \cdot (-\csc^2 x) - \sec^2 x = -3 \cot^2 x \csc^2 x - \sec^2 x \\[10pt]&\text{Differentiate denominator:} \\&\frac{d}{dx} \left[\cos\left(x + \frac{\pi}{4} \right)\right] = -\sin\left(x + \frac{\pi}{4} \right) \\[10pt]&\text{Now evaluate at } x = \frac{\pi}{4}: \\&\quad \cot x = 1 \Rightarrow \cot^2 x = 1 \\&\quad \csc x = \frac{1}{\sin x} = \sqrt{2} \Rightarrow \csc^2 x = 2 \\&\quad \sec x = \frac{1}{\cos x} = \sqrt{2} \Rightarrow \sec^2 x = 2 \\&\quad \sin\left(\frac{\pi}{2} \right) = 1 \\[10pt]&\text{So the expression becomes:} \\&\lim_{x \to \frac{\pi}{4}} \frac{-3(1)(2) - 2}{-1} = \frac{-6 - 2}{-1} = \frac{-8}{-1} = {8}\end{aligned}$$​​