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Q.
If $f(x)=\begin{vmatrix}\sin x & \cos x & \tan x \\x^{3} & x^{2} & x \\2 x & 1 & x
\end{vmatrix}$ then $\displaystyle\lim_{x\to1} \frac {f (x)}{x^2}$ is equal to
$f(x)=\left|\begin{array}{ccc}\sin x & \cos x & \tan x \\ x^{3} & x^{2} & x \\ 2 x & 1 & x\end{array}\right|$
Expand the determinant along the first row,
$\Rightarrow f(x)=\sin x\left(x^{3}-2 x\right)-\cos x\left(x^{4}-2 x^{2}\right)+\tan x\left(x^{3}-2 x^{3}\right)$
$\Rightarrow \frac{ f ( x )}{ x ^{2}}=\frac{\sin x }{ x }\left( x ^{2}-2\right)-\cos x\left( x ^{2}-2\right)- x \tan x$
$\therefore \lim _{x \rightarrow 0} \frac{f(x)}{x^{2}}=\lim _{x \rightarrow 0} \frac{\sin x}{x}\left(x^{2}-2\right)-\lim _{x \rightarrow 0} \cos x\left(x^{2}-2\right)-\lim _{x \rightarrow 0} x \tan x$
$=1(0-2)-1(0-2)-0=0$