Question

If $f(x)=\left|\begin{array}{ccc}\sin x& \cos x& \tan x\\ x^3& x^2& x\\ 2x& 1& x\end{array}\right|$ then $\underset{x\to 1}{\lim }\frac{f(x)}{x^2}$ is equal to

• A. 0
• B. 3
• C. 2
• D. 1

Answer 1

Answer: (A)

$f(x)=\left|\begin{array}{ccc}\sin x& \cos x& \tan x\\ x^3& x^2& x\\ 2x& 1& x\end{array}\right|$
Expand the determinant along the first row,
$\Rightarrow f(x)=\sin x\left(x^3-2x\right)-\cos x\left(x^4-2x^2\right)+\tan x\left(x^3-2x^3\right)$
$\Rightarrow \frac{f(x)}{x^2}=\frac{\sin x}{x}\left(x^2-2\right)-\cos x\left(x^2-2\right)-x\tan x$
$\therefore {\lim }_{x\to 0}\frac{f(x)}{x^2}={\lim }_{x\to 0}\frac{\sin x}{x}\left(x^2-2\right)-{\lim }_{x\to 0}\cos x\left(x^2-2\right)-{\lim }_{x\to 0}x\tan x$
$=1(0-2)-1(0-2)-0=0$