$\mathop {\lim }\limits_{x \to 0^ + } \frac{x e^{1/x}}{1 + e^{1/x}} = $

$\lim _{x \rightarrow-\infty} \frac{3|x|^3-x^2+2|x|-5}{-5|x|^3+3 x^2-2|x|+7} = $

Let $t_{n}$ denote the $n^{th}$ term of the infinite series $\frac{1}{1 !} + \frac{10}{2 !} + \frac{21}{3 !} + \frac{34}{4 !} + \frac{49}{5 !} + \ldots$. Then $\lim _{n \rightarrow \infty} t_{n}$ is

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

If $f(x) = \begin{cases} \frac{\sin([x])}{[x]}, & \text{when } [x] \neq 0 \\ 0, & \text{when } [x] = 0 \end{cases}$ where $[x]$ is the greatest integer function,then $\lim_{x \to 0} f(x) = $

$\mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {1 - \cos 2x} \right)}^2}}}{{2x\tan x - x\tan 2x}}$ is