$\mathop {\lim }\limits_{x \to 0} \frac{{5\sin x + x\cos x}}{{2\tan x - {x^2}}}$ ની કિંમત શોધો.

$\lim_{x \rightarrow 0} \frac{\sin(\pi \sin^2 x)}{x^2} = $

લક્ષ $\lim _{\theta \rightarrow 0} \frac{\tan (\pi \cos ^{2} \theta)}{\sin (2 \pi \sin ^{2} \theta)}$ ની કિંમત શોધો :

$\mathop {\lim }\limits_{x \to 0} \frac{{\sin (\pi {{\cos }^2}x)}}{{{x^2}}} = $

$\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\tan 3x}}{x} + \cos x} \right) = $

$\lim _{x \rightarrow 0} \frac{8}{\sin ^8 x} \left\{1-\cos \left(\frac{x^2}{2}\right)-\cos \left(\frac{x^2}{4}\right)+\cos \left(\frac{x^2}{2}\right) \cos \left(\frac{x^2}{4}\right)\right\} =$