Let $f(2) = 4$ and $f'(2) = 4$,then $\mathop {\lim }\limits_{x \to 2} \,\frac{{xf(2) - 2f(x)}}{{x - 2}}$ equals

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

If $a > 0$ and $\lim _{x \rightarrow a} \frac{a^x - x^a}{x^x - a^a} = -1$,then $a$ is equal to

Let $f: R \rightarrow R$ be a continuous function. Then $\lim _{x \rightarrow \frac{\pi}{4}} \frac{\frac{\pi}{4} \int_{2}^{\sec ^{2} x} f(t) dt}{x^{2}-\frac{\pi^{2}}{16}}$ is equal to :

$\lim _{x \rightarrow 0} \frac{\tan x - \sin x}{x^3}$ is equal to

The value of the limit of $\frac{x^3 - x^2 - 18}{x - 3}$ as $x$ tends to $3$ is