$\mathop {\lim }\limits_{x \to \infty } (\sqrt {{x^2} + 1} - x)$ is equal to

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

$\mathop {\lim }\limits_{x \to {2^ + }} \frac{{1 - \cos \{ {x^2} + 2x\} }}{{\ln {{(x - 1)}^{(x - 2)}}}}$ is equal to (where $\{.\}$ denotes fractional part function).

$\lim _{x \rightarrow \infty}\left(\frac{6 x^2-\cos 3 x}{x^2+5}-\frac{5 x^3+3}{\sqrt{x^6+2}}\right) = $

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

$\lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{2 x^2+x-3} = $