$\mathop {\lim }\limits_{x \to 0} \left( {\frac{{{a^x} - {b^x}}}{x}} \right) = $

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

$\lim _{x \rightarrow 0} \frac{1-\cos \left(x^2+\pi(x+2)\right)}{x^2} = $

If $\mathop {\lim }\limits_{x \to 0} \frac{{\log (a + x) - \log a}}{x} + k\mathop {\lim }\limits_{x \to e} \frac{{\log x - 1}}{{x - e}} = 1$,then

Let $f(x)=5-|x-2|$ and $g(x)=|x+1|$,$x \in R$. If $f(x)$ attains its maximum value at $\alpha$ and $g(x)$ attains its minimum value at $\beta$,then $\lim _{x \rightarrow-\alpha \beta} \frac{(x-1)(x^2-5x+6)}{(x^2-6x+8)}$ is equal to

The value of $\mathop {\lim }\limits_{x \to - 1} \frac{{{x^2} + 3x + 2}}{{{x^2} + 4x + 3}}$ is equal to