Question

Show that the lim(n^2/n!) = 0. Anyone ?

Answer 1

What is n approaching?

Answer 2

infinity

Answer 3

It's like this:

\[\lim_{n \rightarrow \infty}\frac{ n^2 }{ n! }\]
\[=\lim_{n \rightarrow \infty}\frac{ n \times n }{ n \times (n-1)\times (n-2)\times...\times2\times1}\]
\[=\lim_{n \rightarrow \infty}\frac{ n }{ (n-1)\times (n-2)\times...\times2\times1}\]
\[=\lim_{n \rightarrow \infty}\frac{ 1 }{ (\frac{n-1}{n})\times (n-2)\times...\times2\times1}\]
\[=\lim_{n \rightarrow \infty}\frac{ 1 }{ (1-\frac{1}{n})\times (n-2)\times...\times2\times1}\]

You can see now that if you were to substitute n for infinity, the first factor in the denominator would become 1, and the second factor would become infinity. Regardless of how many other factors you have, this makes the entire denominator effectively infinity, and thus makes the fraction equal to 0.

Hope that helps!

Answer 4

Thank you SO much!! Much appreciated! I understand perfectly well from your explanation :)