Question

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

Answer 1

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

$C_2\to \frac{C_2}{x}$

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

$R_2\to R_2-R_3$

$\Rightarrow f\left(x\right)=x\left|\begin{array}{ccc}\cos x& 1& 1\\ \sin x& 0& x\\ \sin x& 1& x\end{array}\right|$

$\Rightarrow x[\cos x(0-x)-1(x\sin x-x\sin x)+1(\sin x-0)]$

$\Rightarrow x[-x\cos x+\sin x]=-x^2\cos x+x\sin x$

$\Rightarrow \frac{f\left(x\right)}{x^2}=\frac{-x^2\cos x+x\sin x}{x^2}=-\cos x+\frac{\sin x}{x}$

$\Rightarrow {lim}_{x\to 0}\frac{f\left(x\right)}{x^2}={lim}_{x\to 0}(-\cos x+\frac{\sin x}{x})$

$\Rightarrow {lim}_{x\to 0}\frac{f\left(x\right)}{x^2}=-1+1=0$