$\lim _{x \rightarrow 0} \frac{(1-\cos 2 x) \sin 5 x}{x^2 \sin 3 x}$ ની કિંમત શોધો.

લક્ષની કિંમત શોધો: $\lim_{x \rightarrow \infty} x \sin \left(\frac{2}{x}\right)$

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

$\mathop {\lim }\limits_{x \to 0} \frac{\sin 2x}{x} = $

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

$\mathop {\lim }\limits_{x \to 0} \frac{{{a^{\sin x}} - 1}}{{{b^{\sin x}} - 1}} = $