If $\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{a}{x} - \frac{4}{{{x^2}}}} \right)^{2x}} = {e^3},$ then $a$ 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

If $a, b$ and $c$ are three distinct real numbers and $\lim _{x \rightarrow \infty} \frac{(b-c) x^2+(c-a) x+(a-b)}{(a-b) x^2+(b-c) x+(c-a)}=\frac{1}{2}$,then $a+2 c=$

$\mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + x)}^{1/x}} - e}}{x}$ equals

If $\sum_{r=1}^{n}(2r-1) = x$,then find the value of $\lim_{n}$ ${\rightarrow \infty} \left[ \frac{1^3}{x^2} + \frac{2^3}{x^2} + \frac{3^3}{x^2} + \ldots + \frac{n^3}{x^2} \right]$.

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