$\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)}} = $

If $[x]$ denotes the greatest integer $\leq x$,then $\lim_{n \rightarrow \infty} \frac{1}{n^3} \{[1^2 x] + [2^2 x] + [3^2 x] + \ldots + [n^2 x] \} = $

If $x_1 = 3$ and $x_{n+1} = \sqrt{2 + x_n}$ for $n \ge 1$,then $\lim_{n \to \infty} x_n$ is equal to

$\mathop {\lim }\limits_{x \to 1} \frac{x - 1}{2x^2 - 7x + 5} = $

$\lim _{x \rightarrow \infty} \frac{(2x^2-3x+5)(3x-1)^{x/2}}{(3x^2+5x+4)\sqrt{(3x+2)^x}}$ is equal to:

The value of $\mathop {\text{Limit}}\limits_{x \to \infty } \frac{{\cot ^{ - 1}}\left( {{x^{ - a}}\log _a x} \right)}{{\sec ^{ - 1}}\left( {{a^x}\log _x a} \right)}$ for $a > 1$ is equal to