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
- A 0
- B 2
- C \(\sqrt{2}\)
- D 1
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. \(\lim _{n \rightarrow \infty}\left(\sqrt{2}-\frac{1}{n}\right)=\sqrt{2}\)
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