$\operatorname{Lim}_{n}$ ${\rightarrow \infty} \left\{ \left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right) \left(2^{\frac{1}{2}}-2^{\frac{1}{5}}\right) \dots \left(2^{\frac{1}{2}}-2^{\frac{1}{2n+1}}\right) \right\}$ is equal to

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

Explore More

Similar Questions

The value of $\lim _{x \rightarrow 0^{+}} \frac{x}{p} \left[ \frac{q}{x} \right]$ is

Let $f: (0, \infty) \rightarrow \mathbb{R}$ and $g: (0, \infty) \rightarrow \mathbb{R}$ be two functions where $g(x) = x + \frac{1}{x}$. If $1 < f(x) \cdot g(x) < 10$ for all $x > 0$,then $\lim_{x \to \infty} f(x)$ is

$\lim _{x \rightarrow \infty}\left(\frac{2+\sin x}{x^2+3}\right)$ is equal to

The value of $\lim _{n \rightarrow \infty} \frac{[r]+[2r]+\ldots+[nr]}{n^{2}}$,where $r$ is a non-zero real number and $[x]$ denotes the greatest integer less than or equal to $x$,is equal to:

$\mathop {\lim }\limits_{x \to 0} \,\left( {\frac{{x - \sin x}}{x}} \right)\,\sin \left( {\frac{1}{x}} \right)$

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo