$\mathop {\lim }\limits_{x \to 0} \frac{{x\tan 2x - 2x\tan x}}{{{{\left( {1 - \cos 2x} \right)}^2}}}$ का मान ज्ञात कीजिए।

$\mathop {\lim }\limits_{\theta \to 0} \frac{{1 - \cos \theta }}{{{\theta ^2}}} = $

मान ज्ञात कीजिए: $\mathop {\lim }\limits_{x \to 0} \frac{\sin 4x}{\sin 2x}$

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

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

$\mathop {\lim }\limits_{x \to 0} \frac{{2{{\sin }^2}3x}}{{{x^2}}} = $