$\lim _{x \rightarrow 0} \frac{x \tan 4x - 2x \tan 2x}{(1 - \cos 4x)^2} = $

$\lim _{x \rightarrow 5} \frac{\sqrt{2-2 \cos \left(x^2-12 x+35\right)}}{(x-5)} = \ldots \ldots$

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

$\lim _{x \rightarrow 0} \frac{(1-\cos 2 x)(3+\cos x)}{x \tan 4 x}$ ની કિંમત શોધો.

$\lim _{x \rightarrow 0} \frac{1-\cos x}{x^2} = $

$\lim _{x}$ ${\rightarrow 0} \frac{\tan \left(\left[-\pi^2\right] x^2\right)-x^2 \tan \left(\left[-\pi^2\right]\right)}{\sin ^2 x}$ ની કિંમત શોધો.