$\mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^{ - 1}}x - {{\tan }^{ - 1}}x}}{{{x^3}}}$ is equal to

$\mathop {\lim }\limits_{\theta \to {0^ + }} {(\sin \theta )^{(\sin \theta - {{\sin }^2}\theta )}}$ is

If $\lim _{t}$ ${\rightarrow 0}\left(\int_0^1(3 x+5)^t d x\right)^{\frac{1}{t}}=\frac{\alpha}{5 e}\left(\frac{8}{5}\right)^{\frac{2}{3}}$,then $\alpha$ is equal to . . . . . .

$\lim _{x}$ ${\rightarrow 0} \frac{x+2 \sin x+3 \tan x-\tan ^3 x}{\sqrt{x^2+2 \sin x+\tan x+3}-\sqrt{\sin ^2 x-2 \tan x-x+3}} =$

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

Let $\alpha$ be a positive real number. Let $f: \mathbb{R} \rightarrow \mathbb{R}$ and $g: (\alpha, \infty) \rightarrow \mathbb{R}$ be the functions defined by $f(x) = \sin \left(\frac{\pi x}{12}\right)$ and $g(x) = \frac{2 \log_{e}(\sqrt{x}-\sqrt{\alpha})}{\log_{e}(e^{\sqrt{x}}-e^{\sqrt{\alpha}})}$. Then the value of $\lim_{x \rightarrow \alpha^{+}} f(g(x))$ is