The value of $\lim _{n \rightarrow \infty} n \sin \frac{2 \pi}{3 n} \cos \frac{2 \pi}{3 n}$ is

Let $[x]$ denote the greatest integer not exceeding $x$. If $l_1 = \lim_{x \rightarrow 2^{+}} (x^2 + [x])$,$l_2 = \lim_{x \rightarrow 3^{-}} (2x - [x])$ and $l_3 = \lim_{x \rightarrow \frac{\pi}{2}} \left( \frac{\cos x}{x - \frac{\pi}{2}} \right)$,then:

If $\lim_{x}$ ${\rightarrow 0} \left\{ \frac{1}{x^{8}} \left( 1 - \cos \frac{x^{2}}{2} - \cos \frac{x^{2}}{4} + \cos \frac{x^{2}}{2} \cos \frac{x^{2}}{4} \right) \right\} = 2^{-k}$,then the value of $k$ is

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

$\lim_{x \rightarrow 0} \frac{x \left( e^{\frac{\sqrt{1+x^{2}+x^{4}}-1}{x}} - 1 \right)}{\sqrt{1+x^{2}+x^{4}}-1}$ is equal to:

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