$ = \cos \left( {{x^2} - 2{x^2}} \right) - x\left( {2\sin x - 2x\tan x} \right)$ $ + 1\left( {2x\sin x - {x^2}\tan x} \right)$
$ = - {x^2}\cos x - 2x\sin x + 2{x^2}\tan x + 2x\sin x - {x^2}\tan x$
$ = {x^2}\tan x - {x^2}\cos x\,\,\,\,\,\, = {x^2}\left( {\tan x - \cos x} \right)$
$ \Rightarrow f'\left( x \right) = 2x\left( {\tan x - \cos x} \right) + {x^2}\left( {{{\sec }^2}x + \sin x} \right)$
$\therefore \mathop {\lim }\limits_{x \to 0} \frac{{f'\left( x \right)}}{x}$
$ = \mathop {\lim }\limits_{x \to 0} \frac{{2x\left( {\tan x - \cos x} \right) + {x^2}\left( {{{\sec }^2}x + \sin x} \right)}}{x}$
$ = \mathop {\lim }\limits_{x \to 0} \left( {\tan x - \cos x} \right) + x\left( {{{\sec }^2}x + \sin x} \right)$
$ = 2\left( {0 - 1} \right) + 0 = - 2$
So, $\mathop {\lim }\limits_{x \to 0} \frac{{f'\left( x \right)}}{x} = - 2$
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