Lt2→∞(2n)!n!nn1n=

Lt2→∞(2n)!n!nn1n = limn→∞[(2n)(2n-1)....(n+1)n!)n!.nn]1n = limn→∞[nn(2)(2-1n)(2-2n)....(1+1n)n!nn.n!]1n =2

Lt2→∞(2n)!n!nn1n=

Lt2→∞(2n)!n!nn1n = limn→∞[(2n)(2n-1)....(n+1)n!)n!.nn]1n = limn→∞[nn(2)(2-1n)(2-2n)....(1+1n)n!nn.n!]1n =2

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$\underset{2 \to \infty}{Lt} {(\frac{(2 n)!}{n! n^{n}})}^{\frac{1}{n}} =$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{2}{e}$

$\frac{3}{e}$

$\frac{4}{e}$

$\frac{1}{e}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\underset{2 \to \infty}{Lt} {(\frac{(2 n)!}{n! n^{n}})}^{\frac{1}{n}} = \lim_{n \to \infty} {[\frac{(2 n) (2 n - 1) . . . . (n + 1) n!)}{n! . n^{n}}]}^{\frac{1}{n}} = \lim_{n \to \infty} {[\frac{n^{n} (2) (2 - \frac{1}{n}) (2 - \frac{2}{n}) . . . . (1 + \frac{1}{n}) n!}{n^{n} . n!}]}^{\frac{1}{n}} = 2$