## anonymous 5 years ago Question on limits

1. anonymous

2. anonymous

$\lim_{n \rightarrow \infty} ((n!)^{1/n})/n$

3. anonymous

fun!

4. anonymous

omg

5. anonymous

Pretty sure the factorial will dominate, but let's find out =)

6. anonymous

givng up

7. anonymous

how do i go about it polpak?

8. anonymous

Well, on top you have $$\infty^0$$

9. anonymous

Which by itself is an indeterminate form

10. anonymous

agreed

11. apples

The limit does exist and is finite.

12. anonymous

yes apples thats correct

13. anonymous

is anyone working on it?

14. apples

I'd suggest using the Squeeze Theorem.

15. anonymous

what is that? would you explain?

16. apples

The squeeze theorem states that, for any function f(x), if you can find g(x) and h(x) such that g(x) <= f(x) <= h(x) and the limit as x goes to a of g(x) is the same as the limit for h(x), then that is also the limit for f(x).

17. apples

Another way to compute the limit would be to use the following definition of e: $e = \lim_{n \to \infty} \frac{n}{\sqrt[n]{n!}}$

18. anonymous

i didnt quite get that theorm.. by using the definition of e it just means that the question i asked yields 1/e as an answer right?

19. apples

Yes.

20. apples

The theorem states that if you can find functions that are upper and lower bounds of your function, and they have the same limit at a certain point, then your function has the same limit at that point.

21. apples

For example, a lower bound for this problem would be $\left(1 - \frac{1}{n}\right)^n$ which has a limit of 1/e as n goes to infinity. So if you could find an upper bound of the problem that also went to 1/e, you could state that the answer to the problem is 1/e by the squeeze theorem.

22. anonymous

ok i understand now. but how is a the lower limit and the upper limit of a function found?

23. apples

You aren't finding "upper and lower" limits, you're finding limits of the upper and lower bounding functions. Let's consider a simple example, f(x) = 0: $\lim_{x \to \infty} f(x)$ An upper bound on this could be something like 1/x, since it is clear that 1/x > 0 for all x > 0. A lower bound could be -1/x, for similar reasons. Since we know that $\lim_{x \to \infty} \frac{1}{x} = \lim_{x \to \infty} \frac{-1}{x} = 0$ and $\frac{-1}{x} <= f(x) <= \frac{1}{x}$ for all x > 0, then $\lim_{x \to \infty} f(x) = 0$

24. anonymous

I think it's easy to find the limit using the definition, that apples just mentioned above, for e.

25. anonymous

The limit we're looking for is just the reciprocal of it. So: $\lim_{n \rightarrow \infty}{n. !^{1 \over n} \over n}=\lim_{n \rightarrow \infty}{1 \over {n \over n.!^{1 \over n}}}={1 \over e}$

26. anonymous

that was a good explanation apples. thank you.

27. anonymous

yea i got it anwar..=)

28. anonymous