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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
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The correct answer is:
$\sqrt{2}$
We have. $n \sqrt{2}-1 < [n \sqrt{2]} \leq n \sqrt{2} \quad[\because x-1 \leq[x] \leq x]$
$\Rightarrow \quad \sqrt{2}-\frac{1}{n} < [n \sqrt{2}] \leq 1$
$\therefore$ By Sandwich theorem. $\lim _{n \rightarrow \infty}\left(\sqrt{2}-\frac{1}{n}\right)=\sqrt{2}$
$\Rightarrow \quad \sqrt{2}-\frac{1}{n} < [n \sqrt{2}] \leq 1$
$\therefore$ By Sandwich theorem. $\lim _{n \rightarrow \infty}\left(\sqrt{2}-\frac{1}{n}\right)=\sqrt{2}$
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