Question

Let $f\left(x\right)=\left|\begin{array}{ccc}\cos x& x& 1\\ 2\sin x& x^2& 2x\\ \tan x& x& 1\end{array}\right|.$ The value of $\underset{x\to 0}{lim}\frac{f\left(x\right)}{x}$ is equal to

• A. $1$
• B. $-1$
• C. $0$
• D. $2$

Answer 1

The correct option is C $0$

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

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

$f^{\prime }\left(x\right)=\left|\begin{array}{ccc}-\sin x& 1& 0\\ 2\sin x& x^2& 2x\\ \tan x& x& 1\end{array}\right|+\left|\begin{array}{ccc}\cos x& x& 1\\ 2\cos x& 2x& 2\\ \tan x& x& 1\end{array}\right|+\left|\begin{array}{ccc}\cos x& x& 1\\ 2\sin x& x^2& 2x\\ {\sec }^{2}x& 1& 0\end{array}\right|$

Thus, $\underset{x\to 0}{lim}\frac{f\left(x\right)}{x}=f^{\prime }\left(0\right)$
$=\left|\begin{array}{ccc}0& 1& 0\\ 0& 0& 0\\ 0& 0& 1\end{array}\right|+\left|\begin{array}{ccc}1& 0& 1\\ 2& 0& 2\\ 0& 0& 1\end{array}\right|+\left|\begin{array}{ccc}1& 0& 1\\ 0& 0& 0\\ 1& 1& 0\end{array}\right|$
$=0$

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

$R_3\to R_3-R_1,$ we get
$f\left(x\right)=\left|\begin{array}{ccc}\cos x& x& 1\\ 2\sin x& x^2& 2x\\ \tan x-\cos x& 0& 0\end{array}\right|$

Expanding along $R_3,$
$f\left(x\right)=(\tan x-\cos x)x^2$
$\underset{x\to 0}{lim}\frac{f\left(x\right)}{x}=\underset{x\to 0}{lim}(\tan x-\cos x)x=0$