$\lim _{x \rightarrow 0} \frac{e^{2|\sin x|}-2|\sin x|-1}{x^2}$

The $\mathop {\lim }\limits_{x \to 0} {(\cos x)^{\cot x}}$ is

If $f(x)$ is a differentiable function and $f''(0) = a$,then $\mathop {\lim }\limits_{x \to 0} \frac{2f(x) - 3f(2x) + f(4x)}{x^2}$ is (in $a$)

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 $f(0) = 2,$ then $\mathop {\lim }\limits_{x \to 0} \frac{{\int\limits_0^x {\left( {tf(x) + xf(t)} \right)dt} }}{{{x^2}}}$ is equal to -

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