$\mathop {\lim }\limits_{x \to a} \frac{{({x^{ - 1}} - {a^{ - 1}})}}{{x - a}} = $

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 differentiable function such that $f \left(\frac{\pi}{4}\right)=\sqrt{2}$,$f \left(\frac{\pi}{2}\right)=0$ and $f^{\prime}\left(\frac{\pi}{2}\right)=1$. Let $g(x)=\int\limits_{x}^{\pi / 4}\left(f^{\prime}(t) \sec t+\tan t \sec t f(t)\right) d t$ for $x \in\left[\frac{\pi}{4}, \frac{\pi}{2}\right)$. Then $\lim\limits _{ x \rightarrow\left(\frac{\pi}{2}\right)^{-}} g ( x )$ is equal to

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

$\mathop {\lim }\limits_{x \to 0} \frac{{{e^{\tan x}} - {e^x}}}{{\tan x - x}} = $

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