Let $f: R \rightarrow R$ be differentiable at $x=0$. If $f(0)=0$ and $f'(0)=2$, then the value of $\lim _{x \rightarrow 0} \frac{1}{x} [f(x)+f(2 x)+f(3 x)+\ldots+f(2015 x)]$ is

$\mathop {\lim }\limits_{x \to 0} \left[ {\frac{{{e^x} - {e^{\sin x}}}}{{x - \sin x}}} \right]$ is equal to

Let $f(x)$ be differentiable at $x = h$. Then $\lim_{x \to h} \frac{(x + h)f(x) - 2hf(h)}{x - h}$ is equal to

Evaluate the limit: $\lim _{x \rightarrow 1} \left[\frac{\sqrt{x}-1}{\log x}\right]$

If $\alpha = \lim_{x \rightarrow \pi/4} \frac{\tan^{3} x - \tan x}{\cos(x + \pi/4)}$ and $\beta = \lim_{x \rightarrow 0} (\cos x)^{\cot x}$ are the roots of the equation $ax^{2} + bx - 4 = 0$,then the ordered pair $(a, b)$ is:

$\mathop {\lim }\limits_{x \to a} \frac{{\cos x - \cos a}}{{\cot x - \cot a}} = $