Question

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

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

Answer 1

The correct option is C

$\sqrt{2}$

Explanation for the correct option.

We know that $x-1\le \left[x\right]\le x$. So, in the given case

$n\sqrt{2}-1\le \left[n\sqrt{2}\right]\le n\sqrt{2}$

Dividing by $n$ we get,

$$\begin{gathered} \frac{n\sqrt{2}-1}{n}\le \frac{\left[n\sqrt{2}\right]}{n}\le \frac{n\sqrt{2}}{n} \\ =\frac{n\sqrt{2}}{n}-\frac{1}{n}\le \frac{\left[n\sqrt{2}\right]}{n}\le \frac{n\sqrt{2}}{n} \\ =\sqrt{2}-\frac{1}{n}\le \frac{\left[n\sqrt{2}\right]}{n}\le \sqrt{2} \end{gathered}$$

By applying limit, we get

$$\begin{gathered} \underset{n\to \infty }{lim}\left(\sqrt{2}-\frac{1}{n}\right)\le \underset{n\to \infty }{lim}\frac{\left[n\sqrt{2}\right]}{n}\le \underset{n\to \infty }{lim}\sqrt{2} \\ =\sqrt{2}-\frac{1}{\infty }\le \underset{n\to \infty }{lim}\frac{\left[n\sqrt{2}\right]}{n}\le \sqrt{2} \\ =\sqrt{2}\le \underset{n\to \infty }{lim}\frac{\left[n\sqrt{2}\right]}{n}\le \sqrt{2} \end{gathered}$$

Therefore, $\underset{n\to \infty }{lim}\frac{\left[n\sqrt{2}\right]}{n}=\sqrt{2}$

Hence, option C is correct.