Q111 / 200 · WBJEE 2014 · Mathematics

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

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

Solution

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. $\lim _{n \rightarrow \infty}\left(\sqrt{2}-\frac{1}{n}\right)=\sqrt{2}$

Hence, the correct answer is option: 'C'.

Relevant Questions