Verified Solution • 2025–26 Session moderate

QJEEM23Ma6379: JEE Main 2023 Mathematics - Class 12 JEEM23Ma6379 Solution: lim n → ∞ { ( 2 1 2 − 2 1 3 ) ( 2 1 2 − 2 1 5 ) … . . ( 2 1 2 − …

limn→∞{(212−213)(212−215)…..(212−212n+1)} is equal to :

Multiple Choice Options

  • A 2\sqrt{2}
  • B 1
  • C 12\frac{1}{\sqrt{2}}
  • D 0

Step-by-Step Verified Solution

Let L=lim⁡n→∞{(212−213)(212−215)…(212−212n−1)}L=\lim\limits_{n \rightarrow \infty}\left\{\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)\left(2^{\frac{1}{2}}-2^{\frac{1}{5}}\right) \ldots\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n-1}}\right)\right\}

Here, \left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)^n<\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)\left(2^{\frac{1}{3}}-2^{\frac{1}{5}}\right) \ldots\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n-1}}\right)<\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n-1}}\right)^n

\Rightarrow \lim\limits_{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)^n < L < \lim\limits_{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n-1}}\right)^n

Since, lim⁡n→∞(212−213)n=0\lim\limits_{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)^n=0 and lim⁡n→∞(212−212n−1)n=0\lim\limits_{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n-1}}\right)^n=0

∴L=0\therefore L = 0

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