The value of the limit $\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - {e^{ - x}} - 2x}}{{x - \sin x}}$ 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 $.40$)

Evaluate $\lim _{x \rightarrow 0^{+}} (x^{n} \ln x)$ for $n > 0$.

If $\alpha = \lim_{x \rightarrow 0} \frac{x \cdot 2^x - x}{1 - \cos x}$ and $\beta = \lim_{x \rightarrow 0} \frac{x \cdot 2^x - x}{\sqrt{1+x^2} - \sqrt{1-x^2}}$,then

The value of $\mathop {\lim }\limits_{x \to 0} \frac{{{{27}^x} - {9^x} - {3^x} + 1}}{{\sqrt 5 - \sqrt {4 + \cos x} }}$ is

$\mathop {Limit}\limits_{x \to 0^+} \frac{1}{x\sqrt{x}} \left( a \tan^{-1} \frac{\sqrt{x}}{a} - b \tan^{-1} \frac{\sqrt{x}}{b} \right)$ has the value equal to