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

The correct option is C $\sqrt{2}$

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

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

Applying $\underset{n\to \infty }{lim}$

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

$\Rightarrow (\sqrt{2}-0)\le \underset{n\to \infty }{lim}\frac{\left[n\sqrt{2}\right]}{n}\le \sqrt{2}(asn\to \infty ;\frac{1}{n}\to 0)$

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