If $\lim _{n \rightarrow \infty} \frac{1-(10)^n}{1+(10)^{n+1}}=\frac{-\alpha}{10}$,then $\alpha$ is equal to

$\mathop {Lim}\limits_{x \to c} f(x)$ does not exist when: (where $[x]$ denotes the greatest integer function and $\{x\}$ denotes the fractional part function.)

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] \} = $

Evaluate the limit: $\lim _{n \rightarrow \infty} \frac{1}{n^3} \sum_{k=1}^n (k^2 x)$

Let $m$ and $n$ be two positive integers greater than $1$. If $\lim_{\alpha \rightarrow 0} \left( \frac{e^{\cos(\alpha^n)} - e}{\alpha^m} \right) = -\left( \frac{e}{2} \right)$,then the value of $\frac{m}{n}$ is

$\lim _{x \rightarrow 8} \frac{\sqrt{1+\sqrt{1+x}}-2}{x-8}$ is equal to