$\lim_{x \to \frac{π}{4}} \frac{\cot^{3} x - \tan x}{\cos x - \sin x} =$

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$4 \sqrt{2}$

$8 \sqrt{2}$

answer is A.

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Detailed Solution

$\begin{matrix} \lim_{x \to \frac{π}{4}} \frac{\cot^{3} x - \tan x}{\cos x - \sin x} = \lim_{x \to \frac{π}{4}} \frac{1 - \tan^{4} x}{\tan^{3} x (\cos x - \sin x)} \\ = \lim_{x \to \frac{π}{4}} \frac{(1 - \tan^{2} x) (1 + \tan^{2} x)}{(\cos x - \sin x)} \\ = \lim_{x \to \frac{π}{4}} \frac{(1 - \tan^{2} x) \sec^{2} x}{(\cos x - \sin x)} \\ = \lim_{x \to \frac{π}{4}} \frac{(1 - \tan^{2} x)}{\cos^{2} x (\cos x - \sin x)} \\ = \lim_{x \to \frac{π}{4}} \frac{(\cos^{2} x - \sin^{2} x)}{\cos^{4} x (\cos x - \sin x)} \\ = \lim_{x \to \frac{π}{4}} \frac{(\cos x + \sin x)}{\cos^{4} x} \\ = 4 \sqrt{2} \end{matrix}$