Question

Question on limits

Answer 1

ask

Answer 2

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

Answer 3

fun!

Answer 4

omg

Answer 5

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

Answer 6

givng up

Answer 7

how do i go about it polpak?

Answer 8

Well, on top you have \(\infty^0\)

Answer 9

Which by itself is an indeterminate form

Answer 10

agreed

Answer 11

The limit does exist and is finite.

Answer 12

yes apples thats correct

Answer 13

is anyone working on it?

Answer 14

I'd suggest using the Squeeze Theorem.

Answer 15

what is that? would you explain?

Answer 16

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).

Answer 17

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

Answer 18

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?

Answer 19

Yes.

Answer 20

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.

Answer 21

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.

Answer 22

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

Answer 23

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 \[\lim_{x \to \infty} f(x) = 0\]

Answer 24

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

Answer 25

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

Answer 26

that was a good explanation apples. thank you.

Answer 27

yea i got it anwar..=)

Answer 28

You already found the answer.. Never mind then :)

Answer 29

No prob