Let $f(x)=\left|\begin{array}{ccc}\cos x & x & 1 \\ 2 \sin x & x & 2 x \\ \sin x & x & x\end{array}\right|$, then $\lim _{x \rightarrow 0} \frac{f(x)}{x^{2}}$ is equal to
(a) 0
(b) $-1$
(c) 2
(d) 3
$f(x)=\left|\begin{array}{ccc}\cos x & x & 1 \\ 2 \sin x & x & 2 x \\ \sin x & x & x\end{array}\right|$
$=\left|\begin{array}{lll}\cos x & x & 1 \\ \sin x & 0 & x \\ \sin x & x & x\end{array}\right|$ $\left[\right.$ Applying $\left.R_{2} \rightarrow R_{2}-R_{3}\right]$
$=\left|\begin{array}{ccc}\cos x & x & 1 \\ \sin x & 0 & x \\ \sin x-\cos x & 0 & x-1\end{array}\right|$ [Applying $R_{3} \rightarrow R_{3}-R_{1}$ ]
$=-x[x \sin x-\sin x-x \sin x+x \cos x]$
$=-x(x \cos x-\sin x)$
$\therefore \lim _{x \rightarrow 0} \frac{f(x)}{x^{2}}=\lim _{x \rightarrow 0} \frac{x(\sin x-x \cos x)}{x^{2}}$
$=\lim _{x \rightarrow 0} \frac{\sin x}{x}-\lim _{x \rightarrow 0} \cos x$
$=1-1=0$
Hence, the correct option is (a).