If $f$ is a real function such that $f(4)=4$ and $f^{\prime}(4)=16$,then $\lim _{x \rightarrow 4} \frac{\sqrt{f(x)}-2}{\sqrt{x}-2} =$

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

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

$\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sqrt{2} \cos x-1}{\cot x-1}$ is equal to

If $f(x) = 3x^{10} - 7x^8 + 5x^6 - 21x^3 + 3x^2 - 7$,then $\lim_{\alpha \rightarrow 0} \frac{f(1-\alpha) - f(1)}{\alpha^3 + 3\alpha} = $