Question

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

• A. $-1$
• B. $0$
• C. $2$
• D. $3$

Answer 1

Answer: (A)

$-1$

For $x\ne 0$

Dividing $R_2$ and $C_2$ by $x$

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

$\therefore {lim}_{x\to 0}\frac{f\left(x\right)}{x^2}=\left|\begin{array}{ccc}1& 1& 1\\ 2\left(1\right)& 1& 2\\ 1& 1& 2\end{array}\right|$

Applying $C_2\to C_2-C_1,C_3\to C_3-C_1$

${lim}_{x\to 0}\frac{f\left(x\right)}{x^2}=\left|\begin{array}{ccc}1& 0& 0\\ 2& -1& 0\\ 1& 0& 1\end{array}\right|=-1$