Question

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

• A. $0$
• B. $2$
• C. $\sqrt{2}$
• D. $1$

Answer 1

Answer: (C)

$\sqrt{2}$

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