$\lim _{x \rightarrow \infty} [x - \log (\cosh x)] = $

$\lim _{x \rightarrow 0} \frac{(\operatorname{cosec} x-\cot x)(e^x-e^{-x})}{\sqrt{3}-\sqrt{2+\cos x}} = $

Let $[x]$ denote the greatest integer not exceeding $x$. If $l_1 = \lim_{x \rightarrow 2^{+}} (x^2 + [x])$,$l_2 = \lim_{x \rightarrow 3^{-}} (2x - [x])$ and $l_3 = \lim_{x \rightarrow \frac{\pi}{2}} \left( \frac{\cos x}{x - \frac{\pi}{2}} \right)$,then:

If $a = \lim_{x \rightarrow 0} \frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2}}{x^4}$ and $b = \lim_{x \rightarrow 0} \frac{\sin^2 x}{\sqrt{2}-\sqrt{1+\cos x}}$,then the value of $ab^3$ is

$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{{1 - {n^2}}} + \frac{2}{{1 - {n^2}}} + \frac{3}{{1 - {n^2}}} + \dots + \frac{n}{{1 - {n^2}}}} \right] =$

If $0 \leq x \leq \pi / 2$,then $\lim _{x \rightarrow a} \frac{|2 \cos x-1|}{2 \cos x-1}$