Question on limits

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

Question on limits

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question