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WBJEE · Maths · Limits

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

  1. A 0
  2. B 2
  3. C \(\sqrt{2}\)
  4. D 1
Verified Solution

Answer & Solution

Correct Answer

(C) \(\sqrt{2}\)

Step-by-step Solution

Detailed explanation

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