Question

If $f\left(x\right)=\left|\begin{array}{ccc}\cos x& x& 1\\ 2\sin x& x^2& 2x\\ \tan x& x& 1\end{array}\right|$, then
$\underset{x\to 0}{lim}\frac{f^{\prime }\left(x\right)}{x}$

• A. does not exist.
• B. exists and is equal to $2$.
• C. exists and is equal to $0$.
• D. exists and is equal to $-2$.

Answer 1

The correct option is D exists and is equal to $-2$.

$f\left(x\right)=\left|\begin{array}{ccc}\cos x& x& 1\\ 2\sin x& x^2& 2x\\ \tan x& x& 1\end{array}\right|$
Using row operations,
$R_1\to R_1-R_3$
$f\left(x\right)=\left|\begin{array}{ccc}\cos x-\tan x& 0& 0\\ 2\sin x& x^2& 2x\\ \tan x& x& 1\end{array}\right|\Rightarrow f\left(x\right)=(\cos x-\tan x)(-x^2)\Rightarrow f\left(x\right)=x^2(\tan x-\cos x)\Rightarrow f^{\prime }\left(x\right)=2x(\tan x-\cos x)+x^2({\sec }^{2}x+\sin x)$
So,
$\underset{x\to 0}{lim}\frac{f^{\prime }\left(x\right)}{x}=\underset{x\to 0}{lim}2(\tan x-\cos x)+x({\sec }^{2}x+\sin x)=-2$