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

$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 - {x^2}} - \sqrt {1 + {x^2}} }}{{{x^2}}}$ is equal to

Let $[t]$ denote the greatest integer $\leq t$. If for some $\lambda \in R - \{0, 1\}$,$\lim_{x \rightarrow 0} \left| \frac{1-x+|x|}{\lambda-x+[x]} \right| = L$,then $L$ is equal to

$\mathop {\lim }\limits_{n \to \infty } \left\{ {\frac{1}{{{n^2}}} + \frac{2}{{{n^2}}} + \frac{3}{{{n^2}}} + \dots + \frac{n}{{{n^2}}}} \right\}$ is

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] \} = $

$\lim _{x \rightarrow 0}(1+3x)^{\frac{2}{x}} = $