$\mathop {\lim }\limits_{x \to \pi /2} \tan x \log \sin x = $

If $f(x)$ is a differentiable function and $f''(0) = a$,then $\mathop {\lim }\limits_{x \to 0} \frac{2f(x) - 3f(2x) + f(4x)}{x^2}$ is (in $a$)

Evaluate the limit: $\lim _{x \rightarrow \pi / 6} \frac{3 \sin x-\sqrt{3} \cos x}{6 x-\pi}$

Value of $\mathop {\lim }\limits_{x \to 1 } \frac{{\left( {\log \left( {1 + x} \right) - \log 2} \right)\left( {3 \cdot 4^{x - 1} - 3x} \right)}}{{\left( {{{\left( {7 + x} \right)}^{1/3}} - {{\left( {1 + 3x} \right)}^{1/2}}} \right)\sin \pi x}}$

$\lim _{x \rightarrow 0} \frac{e^{2|\sin x|}-2|\sin x|-1}{x^2}$

$\mathop {\lim }\limits_{\alpha \to \beta } \left[ {\frac{{{{\sin }^2}\alpha - {{\sin }^2}\beta }}{{{\alpha ^2} - {\beta ^2}}}} \right] = $