$\mathop {\lim }\limits_{x \to 0^+} {x^x} = $

Evaluate the limit: $\lim _{x \rightarrow -9} \frac{(2.5)^{81-x^2}-(0.4)^{x+9}}{x+9}$

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

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

$\lim\limits_{x \rightarrow 0} \frac{\int\limits_{0}^{x} t \sin (10 t) d t}{x}$ is equal to

If $f$ is a strictly increasing function,then $\mathop {\lim }\limits_{x \to 0} \frac{{f({x^2}) - f(x)}}{{f(x) - f(0)}}$ is equal to