If $f\left(x\right)=\begin{vmatrix} x & \sin x & \cos x \\ x^2 & \tan x & -x^3 \\ 2x & \sin 2x & 5x \end{vmatrix}$, $\;$ show that $\lim\limits_{x \to 0} \dfrac{f'\left(x\right)}{x}=-4$
$f'\left(x\right) = \begin{vmatrix}
1 & \sin x & \cos x \\
2x & \tan x & -x^3 \\
2 & \sin 2x & 5x
\end{vmatrix} + \begin{vmatrix}
x & \cos x & \cos x \\
x^2 & \sec^2 x & -x^3 \\
2x & 2\cos 2x & 5x
\end{vmatrix} + \begin{vmatrix}
x & \sin x & -\sin x \\
x^2 & \tan x & -3x^2 \\
2x & \sin 2x & 5
\end{vmatrix}$
$\therefore$ $\;$ $\dfrac{f'\left(x\right)}{x} = \begin{vmatrix}
1 & \sin x & \cos x \\
2 & \tan x / x & -x^2 \\
2 & \sin 2x & 5x
\end{vmatrix} + \begin{vmatrix}
1 & \cos x & \cos x \\
x & \sec^2 x & -x^3 \\
2 & 2 \cos 2x & 5x
\end{vmatrix}+ \begin{vmatrix}
1 & \sin x & -\sin x \\
x & \tan x & -3x^2 \\
2 & \sin 2x & 5
\end{vmatrix}$
$\begin{aligned}
\therefore \; \lim\limits_{x \to 0} \dfrac{f'\left(x\right)}{x} & = \begin{vmatrix}
1 & 0 & 1 \\
2 & 1 & 0 \\
2 & 0 & 0
\end{vmatrix} + \begin{vmatrix}
1 & 1 & 1 \\
0 & 1 & 0 \\
2 & 2 & 0
\end{vmatrix}+ \begin{vmatrix}
1 & 0 & 0 \\
0 & 0 & 0 \\
2 & 0 & 5
\end{vmatrix} \\
& = -2 -2 + 0 \\
& =-4
\end{aligned}$