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. Exists and is equal to $-2$
• B. Does not exist
• C. Exist and is equal to $0$
• D. Exists and is equal to $2$

Answer 1

Answer: (A)

Exists and is equal to $-2$

$f\left(x\right)=\left|\begin{array}{ccc}\cos x& x& 1\\ 2\sin x& x^2& 2x\\ \tan x& x& 1\end{array}\right|$

$=\cos x(x^2-2x^2)-x(2\sin x-2x\tan x)+1(2x\sin x-x^2\tan x)$

$=-x^2\cos x-2x\sin x+2x^2\tan x+2x\sin x-x^2\tan x$

$=x^2\tan x-x^2\cos x$

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

$f^{\prime }\left(x\right)=2x(\tan x-\cos x)+x^2({\sec }^{2}x+\sin x)$

$\therefore \underset{x\to 0}{lim}\frac{f^{\prime }\left(x\right)}{x}=\underset{x\to 0}{lim}\frac{2x(\tan x-\cos x)+x^2({\sec }^{2}x+\sin x)}{x}$

$=\underset{x\to 0}{lim}2(\tan x-\cos x)+x({\sec }^{2}x+\sin x)$

$=2(0-1)+0=-2$

Hence, $\therefore \underset{x\to 0}{lim}\frac{f^{\prime }\left(x\right)}{x}=-2$