$\lim _{x \rightarrow 0} \frac{a^x-1}{\sin (x)} = $

If $\lim_{x \rightarrow 0} [1 + x \ln(1 + b^2)]^{\frac{1}{x}} = 2b \sin^2 \theta$,where $b > 0$ and $\theta \in (-\pi, \pi]$,then the value of $\theta$ is

$\lim _{x \rightarrow 3 / 2} \frac{\left(4 x^2-6 x\right)\left(4 x^2+6 x+9\right)}{\sqrt[3]{2 x}-\sqrt[3]{3}}=$

If $\mathop {\lim }\limits_{x \to 5} \frac{{{x^k} - {5^k}}}{{x - 5}} = 500$,then the positive integral value of $k$ is

$\mathop {Lim}\limits_{x \to {0^ - }} \sin^{-1}([\tan x])$ $= l$,then $\{l\}$ is equal to,where $[\cdot]$ and $\{\cdot\}$ denote the greatest integer and fractional part functions,respectively.

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