$\mathop {\lim }\limits_{x \to 0} \frac{{{e^{\sin x}} - 1}}{x} = $

If $f(9) = 9$ and $f'(9) = 4$,then $\mathop {\lim }\limits_{x \to 9} \frac{{\sqrt {f(x)} - 3}}{{\sqrt x - 3}} = $

Evaluate the limit: $\mathop {\text{Limit}}\limits_{x \to 4} \frac{(\cos \alpha)^x - (\sin \alpha)^x - \cos 2\alpha}{x - 4}$,where $0 < \alpha < \frac{\pi}{2}$.

$\lim\limits_{x \rightarrow 0} \frac{\int\limits_{0}^{x} t \sin (10 t) d t}{x}$ is equal to

$ \lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}} $ is

Let $\alpha$ and $\beta$ be real numbers such that $\lim _{x \rightarrow 0} \frac{1}{x^3}\left(\frac{\alpha}{2} \int_0^x \frac{1}{1-t^2} d t+\beta x \cos x\right)=2$. Then the value of $\alpha+\beta$ is $....$ (in $.40$)