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Q. Let $[x]$ denote the greatest integer less than or equal to $x$ for any real number $x.$ Then $\displaystyle \lim_{n \to \infty}$$\frac{\left[n\sqrt{2}\right]}{n}$ is equal to

WBJEEWBJEE 2014Limits and Derivatives

Solution:

We have,
$n \sqrt{2}-1<[n \sqrt{2]} \leq n \sqrt{2} $
$[\because x-1 \leq[x] \leq x]$
$\Rightarrow \sqrt{2}-\frac{1}{n}<[n \sqrt{2}] \leq 1$
$\therefore $ By Sandwich theorem,
$\displaystyle\lim _{n \rightarrow \infty}\left(\sqrt{2}-\frac{1}{n}\right)=\sqrt{2}$