If \(f(x)=\left|\begin{array}{ccc}\sin x & \cos x & \tan x \\ x^{3} & x^{2} & x \\ 2 x & 1 & x\end{array}\right|\), then
\(\lim _{n -> \infty} \frac{f(x)}{x^{2}}\) is equal to

Given, \(\mathrm{f}(\mathrm{x})=\left|\begin{array}{ccc}\sin \mathrm{x} & \cos x & \tan \mathrm{x} \\ \mathrm{x}^{3} & \mathrm{x}^{2} & \mathrm{x} \\ 2 \mathrm{x} & 1 & \mathrm{x}\end{array}\right|\)
Expanding the determinant along the first row, we get
\[
\begin{aligned}
&=\sin x\left(x^{3}-x\right)-\cos x\left(x^{4}-2 x^{2}\right) \\
&=\sin x\left(x^{3}-x\right)-\cos x \cdot x^{2}\left(x^{2}-2\right) \\
&=\sin x\left(x^{3}-2 x^{3}\right) \\
&\left.\therefore \quad \lim _{x \rightarrow 0} \frac{f(x)}{x^{2}}\right) \tan x\left(-x^{3}\right) \\
&=\lim _{x \rightarrow 0} \frac{\left[x \sin x\left(x^{2}-1\right)-\cos x \cdot x^{2}\left(x^{2}-2\right)\right.}{x^{2}} \cdot x^{2}\left(x^{2}-2\right)-\tan x\left(x^{3}\right) \\
&=\lim _{x \rightarrow 0}\left\{\frac{\sin x}{x}\left(x^{2}-1\right)-\cos x\left(x^{2}-2\right)-\tan x \cdot x\right\} \\
&=[1(0-1)-\cos 0(0-2)-\tan 0 \cdot 0]-[-1+2] \\
&=1
\end{aligned}
\]