Given that $f'(2) = 6$ and $f'(1) = 4$,then $\mathop {\lim }\limits_{h \to 0} \frac{{f(2h + 2 + {h^2}) - f(2)}}{{f(h - {h^2} + 1) - f(1)}} = $

$\mathop {\lim }\limits_{\alpha \to \beta } \left[ {\frac{{{{\sin }^2}\alpha - {{\sin }^2}\beta }}{{{\alpha ^2} - {\beta ^2}}}} \right] = $

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

$A$ 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 \pi /6} \left[ {\frac{{3\sin x - \sqrt 3 \cos x}}{{6x - \pi }}} \right] = $

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