Jeeify
MCQEasyJEE Main 2023 · 6 April, Shift 2Limits

Mathematics Question from JEE Main 2023 · 6 April, Shift 2

lim⁡n→∞{(212−213)(212−215)⋯(212−212n+1)}\lim_{n \to \infty}\left\{\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)\left(2^{\frac{1}{2}}-2^{\frac{1}{5}}\right)\cdots\left(2^{\frac{1}{2}}-2^{\frac{1}{2n+1}}\right)\right\}

is equal to:

  • A

    12\frac{1}{\sqrt{2}}

  • B

    2\sqrt{2}

  • C

    11

  • D

    00

Answer & step-by-step solution

Sign in to reveal the correct answer, the full step-by-step solution, and the common mistakes for this question.

Sign in to see the solution

Practice more Limits questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step - free to start.

Start free trialMore on Limits

Related questions