$\lim _{x \rightarrow 0} \frac{(1-\cos 2 x)}{x \tan 2 x+\frac{2 x}{3} \tan 3 x} = $

$\lim _{x}$ ${\rightarrow 0} \left( \left( \frac{1-\cos ^2(3 x)}{\cos ^3(4 x)} \right) \left( \frac{\sin ^3(4 x)}{(\log _e(2 x+1))^5} \right) \right)$ ની કિંમત $.........$ છે.

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

$\lim _{x \rightarrow 1}(1-x) \tan \left(\frac{\pi}{2} x\right) = $

$\lim _{x \rightarrow 0} \frac{\cos (m x)-\cos (n x)}{x^2} =$

$\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 1}}{{{{\sin }^2}x + \cos x \cos (x + 2) - {{\cos }^2}(x + 1)}}$ ની કિંમત શોધો.