Solve this

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

$=\left|\begin{array}{lll}\cos x & x & 1 \\ \sin x & 0 & x \\ \sin x & x & x\end{array}\right|$     $\left[\right.$ 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).