Let [x] denote the greatest integer less than or equal to x for any real number x. Then displaystyle limn → ∞([n√2]/n) is equal to

Question

Let [x] denote the greatest integer less than or equal to x for any real number x. Then displaystyle limn → ∞([n√2]/n) is equal to (A) 0 (B) 2 (C) √2 (D) 1

Tardigrade Admin · Accepted Answer

We have, n √2-1<[n √2] ≤ n √2 [∵ x-1 ≤[x] ≤ x] ⇒ √2-(1/n)<[n √2] ≤ 1 ∴ By Sandwich theorem, displaystyle lim n arrow ∞(√2-(1/n))=√2

Q. Let $[x]$ denote the greatest integer less than or equal to $x$ for any real number $x .$ Then $n \to \infty lim$ $\frac{[ n 2 ]}{n}$ is equal to

$0$

$2$

$2$

$1$

We have,
$n 2 - 1 < [n 2] \leq n 2$
$[∵ x - 1 \leq [x] \leq x]$
$\Rightarrow 2 - \frac{1}{n} < [n 2] \leq 1$
$∴$ By Sandwich theorem,
$n \to \infty lim (2 - \frac{1}{n}) = 2$

Q. Let [x] denote the greatest integer less than or equal to x for any real number x. Then n→∞lim​n[n2​]​ is equal to

0

2

2​

1

We have, n2​−1<[n2]​≤n2​ [∵x−1≤[x]≤x] ⇒2​−n1​<[n2​]≤1 ∴ By Sandwich theorem, n→∞lim​(2​−n1​)=2​

Q. Let $[x]$ denote the greatest integer less than or equal to $x$ for any real number $x .$ Then $n \to \infty lim$ $\frac{[ n 2 ]}{n}$ is equal to

$0$

$2$

$2$

$1$

We have,
$n 2 - 1 < [n 2] \leq n 2$
$[∵ x - 1 \leq [x] \leq x]$
$\Rightarrow 2 - \frac{1}{n} < [n 2] \leq 1$
$∴$ By Sandwich theorem,
$n \to \infty lim (2 - \frac{1}{n}) = 2$