$$\lim_{x \to \frac{\pi}{4}} \frac{\cot^3 x - \tan x}{\cos\left(x + \frac{\pi}{4}\right)}$$ is

We have to evaluate the limit

$$\displaystyle L=\lim_{x\to\frac{\pi}{4}} \frac{\cot^{3}x-\tan x}{\cos\!\left(x+\dfrac{\pi}{4}\right)}.$$

First we note the values of the functions at the point of approach. When $$x=\dfrac{\pi}{4}$$ we get $$\tan\dfrac{\pi}{4}=1,\qquad \cot\dfrac{\pi}{4}=1,$$ hence

$$\cot^{3}\dfrac{\pi}{4}-\tan\dfrac{\pi}{4}=1^{3}-1=0.$$

Also

$$\cos\!\left(\dfrac{\pi}{4}+\dfrac{\pi}{4}\right) =\cos\dfrac{\pi}{2}=0.$$

Because both the numerator and the denominator approach $$0,$$ the limit is of the indeterminate form $$\dfrac{0}{0}.$$ Therefore we can apply L’Hôpital’s Rule, which says:

$$\text{If } \displaystyle\lim_{x\to a}\frac{f(x)}{g(x)}= \frac{0}{0}\text{ or }\frac{\infty}{\infty}, \text{ then } \displaystyle\lim_{x\to a}\frac{f(x)}{g(x)}= \displaystyle\lim_{x\to a}\frac{f'(x)}{g'(x)},$$ provided the latter limit exists.

So we differentiate the numerator and the denominator with respect to $$x.$$

For the numerator $$f(x)=\cot^{3}x-\tan x$$ we write

$$f'(x)=\frac{d}{dx}\bigl(\cot^{3}x\bigr)-\frac{d}{dx}\bigl(\tan x\bigr).$$

We recall the derivatives $$\frac{d}{dx}(\cot x)=-\csc^{2}x \quad\text{and}\quad \frac{d}{dx}(\tan x)=\sec^{2}x.$$

Using the chain rule on the first term we get

$$\frac{d}{dx}\bigl(\cot^{3}x\bigr)=3\cot^{2}x\cdot(\!-\csc^{2}x) =-3\cot^{2}x\,\csc^{2}x.$$

Thus

$$f'(x)=-3\cot^{2}x\,\csc^{2}x-\sec^{2}x.$$

For the denominator $$g(x)=\cos\!\left(x+\dfrac{\pi}{4}\right)$$ we use the derivative $$\dfrac{d}{dx}\cos u=-\sin u\cdot\dfrac{du}{dx}.$$ Here $$u=x+\dfrac{\pi}{4},$$ so $$\dfrac{du}{dx}=1,$$ and consequently

$$g'(x)=-\sin\!\left(x+\dfrac{\pi}{4}\right).$$

Now we evaluate the derivatives at $$x=\dfrac{\pi}{4}.$$

First compute the trigonometric values:

$$\cot\dfrac{\pi}{4}=1,\; \csc\dfrac{\pi}{4}=\frac{1}{\sin\dfrac{\pi}{4}} =\frac{1}{\dfrac{\sqrt2}{2}}=\sqrt2,\; \sec\dfrac{\pi}{4}=\frac{1}{\cos\dfrac{\pi}{4}} =\frac{1}{\dfrac{\sqrt2}{2}}=\sqrt2.$$

Hence

$$\cot^{2}\dfrac{\pi}{4}=1,\qquad \csc^{2}\dfrac{\pi}{4}=2,\qquad \sec^{2}\dfrac{\pi}{4}=2.$$

Substituting in $$f'(x)$$ gives

$$f'\!\left(\dfrac{\pi}{4}\right) =-3(1)(2)-2=-6-2=-8.$$

For the denominator we find

$$g'\!\left(\dfrac{\pi}{4}\right) =-\sin\!\left(\dfrac{\pi}{4}+\dfrac{\pi}{4}\right) =-\sin\dfrac{\pi}{2}=-1.$$

Applying L’Hôpital’s Rule we have

$$L=\lim_{x\to\frac{\pi}{4}}\frac{f(x)}{g(x)} =\lim_{x\to\frac{\pi}{4}}\frac{f'(x)}{g'(x)} =\frac{f'\!\left(\dfrac{\pi}{4}\right)} {g'\!\left(\dfrac{\pi}{4}\right)} =\frac{-8}{-1}=8.$$

Hence, the correct answer is Option D.