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

$\lim _{n \rightarrow \infty} \frac{2^2+4^2+6^2+\ldots+(2 n)^2}{n^3} = $

Let $a_1, a_2, a_3, \ldots, a_n$ be $n$ positive consecutive terms of an arithmetic progression. If $d > 0$ is its common difference,then evaluate $\lim_{n \rightarrow \infty} \sqrt{\frac{d}{n}} \left( \frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + \ldots + \frac{1}{\sqrt{a_{n-1}} + \sqrt{a_n}} \right)$.

$\lim _{x \rightarrow \infty} \left(\frac{x+a}{x+b}\right)^{x}$ is equal to

Find the limit: $\mathop {\lim }\limits_{x \to 2} \left[\frac{x^{2}-4}{x^{3}-4 x^{2}+4 x}\right]$

If $f(x) = \begin{cases} \frac{\sin(1+[x])}{[x]}, & \text{for } [x] \neq 0 \\ 0, & \text{for } [x] = 0 \end{cases}$ where $[x]$ denotes the greatest integer function,then $\lim_{x \rightarrow 0^{-}} f(x)$ is equal to