$\mathop {\lim }\limits_{x \to 0} x \log (\sin x) = $

Evaluate $\mathop {\lim }\limits_{x \to \infty } \frac{{\log {x^n} - [x]}}{{[x]}},$ where $n \in N$ and $[x]$ denotes the greatest integer less than or equal to $x$.

Evaluate the limit: $\mathop {\text{Limit}}\limits_{x \to 4} \frac{(\cos \alpha)^x - (\sin \alpha)^x - \cos 2\alpha}{x - 4}$,where $0 < \alpha < \frac{\pi}{2}$.

The limit of $\left\{\frac{1}{x} \sqrt{1+x}-\sqrt{1+\frac{1}{x^{2}}}\right\}$ as $x \rightarrow 0$ is:

$\mathop {\lim }\limits_{x \to \pi /2} \left[ {x\tan x - \left( {\frac{\pi }{2}} \right)\sec x} \right] = $

Let $f(x) = \lim_{n \rightarrow \infty} \sum_{r=0}^n \left( \frac{2\tan(x/2^{r+1})}{1 - \tan^2(x/2^{r+1})} \right)$. Then $\lim_{x \rightarrow 0} \frac{e^x - e^{f(x)}}{x - f(x)}$ is equal to . . . . . . .