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

Let $f: R \to R$ be a positive increasing function with $\lim_{x \to \infty} \frac{f(3x)}{f(x)} = 1$. Then $\lim_{x \to \infty} \frac{f(2x)}{f(x)} = $

The value of $\lim _{n \rightarrow \infty} \frac{[r]+[2r]+\ldots+[nr]}{n^{2}}$,where $r$ is a non-zero real number and $[x]$ denotes the greatest integer less than or equal to $x$,is equal to:

If $\lim _{x \rightarrow 0} \frac{|x|}{\sqrt{x^4+4 x^2+5}}=k$ and $\lim _{x \rightarrow 0} x^4 \sin \left(\frac{1}{3 \sqrt{x}}\right)=l$,then $k+l=$

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

$\mathop {\lim }\limits_{x \to 0} \,\left( {\frac{{x - \sin x}}{x}} \right)\,\sin \left( {\frac{1}{x}} \right)$