$\mathop {\lim }\limits_{x \to 0} \left( \frac{\sin x - x + \frac{x^3}{6}}{x^5} \right) = $

Evaluate the limit: $\lim _{x \rightarrow 0} \frac{e^x-e^{\sin x}}{2(x-\sin x)}$

$\lim _{x \rightarrow 0} \left[ \frac{1}{x} - \frac{1}{e^x - 1} \right] = $

$\lim _{x \rightarrow 1} \frac{\tan \left(x^{2}-1\right)}{x-1}$ is equal to

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

If $f(9)=9$ and $f^{\prime}(9)=4$,then $\lim _{x \rightarrow 9} \frac{\sqrt{f(x)}-3}{\sqrt{x}-3} = $