If $f(x)=\left|\begin{array}{ccc}\cos x & x & 1 \\ 2 \sin x & x^2 & 2 x \\ \tan x & x & 1\end{array}\right|$…
If $f(x)=\left|\begin{array}{ccc}\cos x & x & 1 \\ 2 \sin x & x^2 & 2 x \\ \tan x & x & 1\end{array}\right|$, then $\lim _{x \rightarrow 0} \frac{f^{\prime}(x)}{x}$
Exists and is equal to $-2$
Does not exist
Exist and is equal to 0
Exists and is equal to 2
Solution
$\begin{aligned} & f(x)=\left|\begin{array}{ccc}\cos x & x & 1 \\ 2 \sin x & x^2 & 2 x \\ \tan x & x & 1\end{array}\right| \\=& \cos x\left(x^2-2 x^2\right)-x(2 \sin x-2 x \tan x) \\=&-x^2 \cos x-2 x \sin x+2 x^2 \tan x \\\left.+2 x \sin x-x^2 \tan x\right) \\=& x^2 \tan x-x^2 \tan x \\ \Rightarrow & f^{\prime}(x)=2 x(\tan x-\cos x)+x^2\left(\sec ^2 x+\sin x\right) \\ \therefore & \lim _{x \rightarrow 0} \frac{f^{\prime}(x)}{x} \\=& \lim _{x \rightarrow 0} \frac{2 x(\tan x-\cos x)+x^2\left(\sec ^2 x+\sin x\right)}{x} \\=& \lim _{x \rightarrow 0}(\tan x-\cos x)+x\left(\sec ^2 x+\sin x\right) \\=& 2(0-1)+0=-2 \\ \text { So, } \lim _{x \rightarrow 0} \frac{f^{\prime}(x)}{x}=-2 \end{aligned}$