If $$f(x) = \begin{vmatrix} \cos x & x & 1 \\ 2\sin x & x^2 & 2x \\ \tan x & x & 1 \end{vmatrix}$$, then $$\lim_{x \to 0} \frac{f'(x)}{x}$$:

We have the function

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

To evaluate the required limit, we first simplify this determinant. Using the standard expansion along the first row, namely

$$\begin{vmatrix}a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}\end{vmatrix} =a_{11}\begin{vmatrix}a_{22}&a_{23}\\ a_{32}&a_{33}\end{vmatrix} -a_{12}\begin{vmatrix}a_{21}&a_{23}\\ a_{31}&a_{33}\end{vmatrix} +a_{13}\begin{vmatrix}a_{21}&a_{22}\\ a_{31}&a_{32}\end{vmatrix},$$

we write

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

Each 2 × 2 minor is evaluated with the rule $$\begin{vmatrix}p&q\\ r&s\end{vmatrix}=ps-qr.$$ Hence

$$\begin{aligned} f(x)=\;&\cos x\,[x^{2}\cdot1-(2x)\cdot x] \\ &-x\,[2\sin x\cdot1-(2x)\cdot\tan x] \\ &+1\,[2\sin x\cdot x-x^{2}\cdot\tan x]. \end{aligned}$$

Multiplying out each bracket we get

$$\begin{aligned} f(x)=&\;\cos x\,(x^{2}-2x^{2}) \\ &-x\,(2\sin x-2x\tan x) \\ &amp+\,(2x\sin x-x^{2}\tan x). \end{aligned}$$

Simplify term by term:

$$\cos x\,(x^{2}-2x^{2})=\cos x\,(-x^{2})=-x^{2}\cos x,$$

$$-x\,(2\sin x-2x\tan x)=-2x\sin x+2x^{2}\tan x,$$

$$2x\sin x-x^{2}\tan x$$

Putting them together,

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

Factorising out the common factor $$x^{2},$$ we arrive at the compact form

$$f(x)=x^{2}\bigl(\tan x-\cos x\bigr).$$

Let us denote $$g(x)=\tan x-\cos x,$$ so that $$f(x)=x^{2}g(x).$$

Next, we need the derivative of $$f(x).$$ We first recall the product rule of differentiation: for any functions $$u(x),v(x)$$ we have $$(uv)'=u'v+uv'.$$ Applying it to $$u=x^{2}$$ and $$v=g(x),$$ we obtain

$$f'(x)=\bigl(x^{2}\bigr)'g(x)+x^{2}g'(x)=2x\,g(x)+x^{2}g'(x).$$

The problem asks for the limit

$$\lim_{x\to0}\frac{f'(x)}{x} =\lim_{x\to0}\Bigl(2\,g(x)+x\,g'(x)\Bigr).$$

We therefore evaluate $$g(x)$$ and $$g'(x)$$ at $$x=0.$$ First,

$$g(0)=\tan0-\cos0=0-1=-1.$$

Next, we differentiate $$g(x).$$ Using the known derivatives $$\dfrac{d}{dx}(\tan x)=\sec^{2}x$$ and $$\dfrac{d}{dx}(-\cos x)=\sin x,$$ we get

$$g'(x)=\sec^{2}x+\sin x.$$

Hence at $$x=0$$

$$g'(0)=\sec^{2}0+\sin0=1+0=1.$$

Now substitute these values back into the expression for the limit:

$$\lim_{x\to0}\frac{f'(x)}{x} =\lim_{x\to0}\Bigl(2\,g(x)+x\,g'(x)\Bigr) =2\,g(0)+0\cdot g'(0)=2(-1)+0=-2.$$

Thus the limit exists and equals $$-2.$$ Hence, the correct answer is Option A.