Evaluate the given limit: $\mathop {\lim }\limits_{x \to 0} (\text{cosec}\,x - \cot x)$

If $a = \lim_{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{n^2}$ and $b = \lim_{n \rightarrow \infty} \frac{1^2+2^2+3^2+\ldots+n^2}{n^3}$,then

$\lim _{x \rightarrow 0} \frac{(3^{2x}-\sqrt{x+1}) \sin 5x}{1-\cos 4x} =$

$\lim _{x \rightarrow 0} \frac{63^x-9^x-7^x+1}{\sqrt{2}-\sqrt{1+\cos x}}=\ldots$.

$\mathop {\lim }\limits_{x \to 0} \frac{{\sin ({x^{1/3}})\ln (1 + 3x)}}{{{{(\tan^{ - 1}\sqrt x )}^2}({e^{5{x^{1/3}}}} - 1)}} = $

$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{{{n^3} + 1}} + \frac{4}{{{n^3} + 1}} + \frac{9}{{{n^3} + 1}} + \dots + \frac{{{n^2}}}{{{n^3} + 1}}} \right] = $