Question

Limits

Answer 1

I don't like them too.

Answer 2

\[\LARGE \lim_{n \rightarrow \infty} (\frac{n!}{n^n})^{\frac{1}{n}}\]

Answer 3

I got one word for you guys... SQUEEZE

Answer 4

:O

Answer 5

although I'm a bit shaky with this, would you guys confirm?
\[\huge \left(\frac{1}{n^n}\right)^{\frac1n}\le \left(\frac{n!}{n^n}\right)^{\frac1n}\le \left(\frac{n^n}{n^n}\right)^{\frac1n}\]

Since clearly (?)
\[\huge n^n >n!\]

Answer 6

On second thought, pass.

Answer 7

got a little overenthusiastic :)

Answer 8

\[
\text{for factorials, start by checking with Stirling's approximation: }\\
n!\approx\sqrt{2\pi n}\left(n\over e\right)^n\\
L=\lim_{n\to\infty}\left[{\sqrt{2\pi n}\over n^n}\left(n\over e\right)^n\right]^{1/n}\\
\vdots
\text{simplifying,}\\
\boxed{L={1\over e}}
\]

Answer 9

what you wrote is beyond my scope of understanding :P

Answer 10

All I did was replace n! with its sterling's approximation like I showed

then I left the intermediate steps for you to ejoy and gave you the final answeer

Answer 11

I haven't heard that sterling's approximation :|

Answer 12

whenever you have to play with factorials, thats the best approach.
for example, even for logarithms of factorials, etc..

Answer 13

http://en.wikipedia.org/wiki/Stirling's_approximation

Answer 14

hmm..seems cool :O thanks..you knowledge base is quite high man :|

Answer 15

:) yw.

Answer 16

just an ordinary Joe

Answer 17

^understatement.

Answer 18

got it yet?

Answer 19

simplify the terms in the square paranthesis

Answer 20

or you can remember this as a rule:
\[
\lim_{n\to\infty}{n\over\sqrt[n]{n!}}=e
\]