$\mathop {\lim }\limits_{n \to \infty } \frac{{\sqrt n }}{{\sqrt n + \sqrt {n + 1} }} = $

Let $a_1, a_2, a_3, \ldots, a_n$ be $n$ positive consecutive terms of an arithmetic progression. If $d > 0$ is its common difference,then evaluate $\lim_{n \rightarrow \infty} \sqrt{\frac{d}{n}} \left( \frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + \ldots + \frac{1}{\sqrt{a_{n-1}} + \sqrt{a_n}} \right)$.

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

Evaluate the limit: $\lim _{x}$ ${\rightarrow 0}\left(\frac{4 !}{x^8}\left(1-\cos \frac{x^2}{3}-\cos \frac{x^2}{4}+\cos \frac{x^2}{3} \cos \frac{x^2}{4}\right)\right)$

$\lim _{x \rightarrow 1} \left( \lim _{y \rightarrow \infty} y \left( (e^x)^{1/y} - 1 \right) \right) = $

If $f(x) = \frac{\sin(e^{x-2} - 1)}{\log(x-1)},$ then $\lim_{x \to 2} f(x)$ is given by