Solving a 'Harvard' University entrance exam question

[00:00:00]Hello Friends ∞!/∞^∞=?

[00:00:09]let's have a solution as we know that n!=1.2.3...n n^n=n.n.n...n n times then for any positive 'n' we have n!/n^n will be greater than and equal to zero let, this is first equation also n!/n^n=1.2.3...n/n.n.n...n It will be 1/n.2/n.3/n...n/n since 2/n is less than and equal to 1 3/n is less than and equal to 1 then n/n will also less than and equal to 1 so n!/n^n=1/n let, this is second equation from first equation and second equation 0≤n!/n^n≤1/n we can put limit as n approaches to ∞ Lim n→∞ 0 ≤ Lim n→∞ n!/n^n ≤ Lim n→∞ 1/n 0 ≤ Lim n→∞ (n!/n^n) ≤ 1/∞ as we know 1/∞=0 Lim n→∞ (n!/n^n)=0 so finally ∞!/∞^∞=0 which is our the final answer thanks for watching this video please subscribe this channel to get the notification of my new videos and don't forget to share these videos with your classmates and friends so that they also have a benefit of it ok bye