1. Write the determinant

We are given

$$f(x)=\begin{vmatrix} \cos x & x & 1\\ 2\sin x & x^2 & 2x\\ \tan x & x & 1 \end{vmatrix}.$$

We need

$$\lim_{x\to 0}\frac{f'(x)}{x}.$$

2. Simplify the determinant

Notice that the third column is

$$\begin{pmatrix}1\\2x\\1\end{pmatrix}.$$

Expand along the third column:

$$f(x)=1\cdot \begin{vmatrix}2\sin x & x^2\\ \tan x & x\end{vmatrix} -2x\cdot \begin{vmatrix}\cos x & x\\ \tan x & x\end{vmatrix} +1\cdot \begin{vmatrix}\cos x & x\\ 2\sin x & x^2\end{vmatrix}.$$

Now compute each minor:

$$\begin{vmatrix}2\sin x & x^2\\ \tan x & x\end{vmatrix} =2x\sin x-x^2\tan x,$$ $$\begin{vmatrix}\cos x & x\\ \tan x & x\end{vmatrix} =x\cos x-x\tan x=x(\cos x-\tan x),$$ $$\begin{vmatrix}\cos x & x\\ 2\sin x & x^2\end{vmatrix} =x^2\cos x-2x\sin x.$$

Hence

$$f(x)=(2x\sin x-x^2\tan x)-2x\cdot x(\cos x-\tan x)+(x^2\cos x-2x\sin x).$$

Simplify:

$$f(x)=2x\sin x-x^2\tan x-2x^2\cos x+2x^2\tan x+x^2\cos x-2x\sin x.$$

The $2x\sin x$ terms cancel, so

$$f(x)=x^2(\tan x-\cos x).$$

3. Differentiate

Using product rule,

$$f'(x)=2x(\tan x-\cos x)+x^2(\sec^2 x+\sin x).$$

Therefore

$$\frac{f'(x)}{x}=2(\tan x-\cos x)+x(\sec^2 x+\sin x).$$

4. Take the limit as $x\to 0$

As $x\to 0$,

$$\tan x\to 0,\qquad \cos x\to 1,\qquad x\sec^2 x\to 0,\qquad x\sin x\to 0.$$

So

$$\lim_{x\to 0}\frac{f'(x)}{x}=2(0-1)+0=-2.$$

5. Check options

• A: does not exist — false
• B: equal to $2$ — false
• C: equal to $0$ — false
• D: equal to $-2$ — true

Therefore the correct option is D.