Let $f(x) = \frac{\sqrt{x+3}}{x+1}$. Then the value of $\lim_{x \rightarrow -3^{-}} f(x)$ is

For each $x \in \mathbb{R}$,let $[x]$ represent the greatest integer function. Then $\lim _{x \rightarrow 0^{-}} \frac{x([x]+|x|) \sin [x]}{|x|}$ is equal to

$\lim _{n}$ ${\rightarrow \infty} \frac{\left(1^2-1\right)(n-1)+\left(2^2-2\right)(n-2)+\ldots +\left((n-1)^2-(n-1)\right) \cdot 1}{\left(1^3+2^3+\ldots +n^3\right)-\left(1^2+2^2+\ldots +n^2\right)}$ is equal to:

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

$\lim _{x \rightarrow 2} \frac{3^x+3^{3-x}-12}{3^{3-x}-3^{\frac{x}{2}}} = $

$\mathop {\lim }\limits_{x \to \infty } \frac{{{{(2x + 1)}^{40}}{{(4x - 1)}^5}}}{{{{(2x + 3)}^{45}}}} = $