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anonymous
 5 years ago
Question on limits
anonymous
 5 years ago
Question on limits

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{n \rightarrow \infty} ((n!)^{1/n})/n\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Pretty sure the factorial will dominate, but let's find out =)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how do i go about it polpak?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, on top you have \(\infty^0\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Which by itself is an indeterminate form

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The limit does exist and is finite.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes apples thats correct

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is anyone working on it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'd suggest using the Squeeze Theorem.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what is that? would you explain?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 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).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Another 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
 5 years ago
Best ResponseYou've already chosen the best response.0i 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0For 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
 5 years ago
Best ResponseYou've already chosen the best response.0ok i understand now. but how is a the lower limit and the upper limit of a function found?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You 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
 5 years ago
Best ResponseYou've already chosen the best response.0I think it's easy to find the limit using the definition, that apples just mentioned above, for e.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 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
 5 years ago
Best ResponseYou've already chosen the best response.0that was a good explanation apples. thank you.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea i got it anwar..=)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You already found the answer.. Never mind then :)
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