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

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

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 $.40$)

$A$ weight hangs by a spring and is caused to vibrate by a sinusoidal force. Its displacement $s(t)$ at time $t$ is given by an equation of the form $s(t) = \frac{A}{c^2 - k^2} (\sin kt - \sin ct)$,where $A, c,$ and $k$ are positive constants with $c \neq k$. Then the limiting value of the displacement as $c \to k$ is:

$\mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + x)}^{1/2}} - {{(1 - x)}^{1/2}}}}{x} = $

$\mathop {\lim }\limits_{x \to 0} \frac{{x{e^x} - \log (1 + x)}}{{{x^2}}}$ equals