$\lim _{x \rightarrow \infty}\left(\frac{2+\sin x}{x^2+3}\right)$ is equal to

$\lim _{x \rightarrow 0} x^2 \sin \left(\frac{\pi}{x}\right)$ is equal to

$\lim _{x \rightarrow 2}\left[\left(x^2-4 x+4\right) \cos \left(\frac{2}{x-2}\right)+\frac{x^2-4}{x^3-2 x-4}\right]=$

$\operatorname{Lim}_{n}$ ${\rightarrow \infty} \left\{ \left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right) \left(2^{\frac{1}{2}}-2^{\frac{1}{5}}\right) \dots \left(2^{\frac{1}{2}}-2^{\frac{1}{2n+1}}\right) \right\}$ is equal to

$\lim_{x \rightarrow \infty} \frac{[2x - 3]}{x} = $

Let $[x]$ denote the greatest integer less than or equal to $x$ for any real number $x$. Then,$\lim _{n \rightarrow \infty} \frac{[n \sqrt{2}]}{n}$ is equal to