$\mathop {\lim }\limits_{x \to \frac{\pi }{2}} (1 - \sin x)\tan x$ is

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

The value of $\lim _{x \rightarrow 0} \frac{1}{x^3} \int_0^x \frac{t \ln (1+t)}{t^4+4} dt$ is

$\lim _{x \rightarrow 0}(\sin x)^{2 \tan x}$ 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 :

If $f(0) = 2,$ then $\mathop {\lim }\limits_{x \to 0} \frac{{\int\limits_0^x {\left( {tf(x) + xf(t)} \right)dt} }}{{{x^2}}}$ is equal to -