$\mathop {\lim }\limits_{x \to 0} \frac{{\sin x + \log (1 - x)}}{{{x^2}}}$ is equal to

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

$\lim _{x \rightarrow 1} \frac{a b^x-a^x b}{x^2-1} = $

If $\alpha, \beta$ are the distinct roots of $x^{2}+bx+c=0$,then $\lim _{x \rightarrow \beta} \frac{e^{2(x^{2}+bx+c)}-1-2(x^{2}+bx+c)}{(x-\beta)^{2}}$ is equal to:

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

$\mathop {\lim }\limits_{x \to 0} \left[ {\frac{1}{x} - \frac{{\log (1 + x)}}{{{x^2}}}} \right] =$