anonymous
  • anonymous
Question on limits
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
ask
anonymous
  • anonymous
\[\lim_{n \rightarrow \infty} ((n!)^{1/n})/n\]
anonymous
  • anonymous
fun!

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anonymous
  • anonymous
omg
anonymous
  • anonymous
Pretty sure the factorial will dominate, but let's find out =)
anonymous
  • anonymous
givng up
anonymous
  • anonymous
how do i go about it polpak?
anonymous
  • anonymous
Well, on top you have \(\infty^0\)
anonymous
  • anonymous
Which by itself is an indeterminate form
anonymous
  • anonymous
agreed
apples
  • apples
The limit does exist and is finite.
anonymous
  • anonymous
yes apples thats correct
anonymous
  • anonymous
is anyone working on it?
apples
  • apples
I'd suggest using the Squeeze Theorem.
anonymous
  • anonymous
what is that? would you explain?
apples
  • 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).
apples
  • apples
Another way to compute the limit would be to use the following definition of e: \[e = \lim_{n \to \infty} \frac{n}{\sqrt[n]{n!}}\]
anonymous
  • 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?
apples
  • apples
Yes.
apples
  • 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.
apples
  • 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.
anonymous
  • anonymous
ok i understand now. but how is a the lower limit and the upper limit of a function found?
apples
  • 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 \[\lim_{x \to \infty} f(x) = 0\]
anonymous
  • anonymous
I think it's easy to find the limit using the definition, that apples just mentioned above, for e.
anonymous
  • anonymous
The limit we're looking for is just the reciprocal of it. So: \[\lim_{n \rightarrow \infty}{n. !^{1 \over n} \over n}=\lim_{n \rightarrow \infty}{1 \over {n \over n.!^{1 \over n}}}={1 \over e}\]
anonymous
  • anonymous
that was a good explanation apples. thank you.
anonymous
  • anonymous
yea i got it anwar..=)
anonymous
  • anonymous
You already found the answer.. Never mind then :)
apples
  • apples
No prob

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