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Limit questions

  1. Oct 24, 2005 #1
    find the [tex]\lim_{n\rightarrow\infty}(\frac{n+1}{n})^{n+1} = [/tex]
    so far what i have is
    [tex]\lim_{n\rightarrow\infty}(\frac{n+1}{n})^{n+1} = \lim_{n\rightarrow\infty}(\frac{n}{n} + \frac{1}{n})^{n+1} = \lim_{n\rightarrow\infty}(1 + \frac{1}{n})^{n+1}\\[/tex]

    I know this has got to go to [tex]e[/tex] or something very close, my question lies in if I should be concerned with the +1 in the n+1 in the exponet.
    There is another one which I'm totaly stumped on for a starting point, my gut tells me the limit is 1 or it may not even exist altogether.
    [tex]\lim_{n\rightarrow\infty}(\frac{n+3}{n+1})^{n+4} = [/tex]
    but then again it could be +[tex]\infty[/tex] if you realize that in the () the numerator is larger than the denominator such that as the exponet is applied it goes to infinity instead od zero.
    Last edited: Oct 24, 2005
  2. jcsd
  3. Oct 24, 2005 #2


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    You said you already have a suspicion that the limit should have to do with e. So I assume you know the limit definition of e which has a very similar form?

    The second has a very similar form. Can you rewrite it so the form resembles a limit you already know? (Like that of exp(x) or something).
  4. Oct 24, 2005 #3


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    Did you notice that
    is the same as
    [tex]\left(1+\frac{1}{n}\right)^n \left(1+ \frac{1}{n}\right)[/tex]
  5. Oct 24, 2005 #4
    no I did not realize that [tex]\left(1+\frac{1}{n}\right)^n \left(1+ \frac{1}{n}\right)[/tex] I guess that would make the limit equal to e^2?

    as far as the second problem I don't know what exp(x) is but i recognized that I can get things to look more like e

    [tex]\lim_{n\rightarrow\infty}(\frac{n+3}{n+1})^{n+4} = [/tex]

    [tex]\lim_{n\rightarrow\infty}((\frac{n + 3}{n + 1})(\frac{1/n}{1/n}))^{n+4} = [/tex]
    [tex]\lim_{n\rightarrow\infty}\frac{(1 + (3/n))^{n+4}}{(1 + (1/n))^{n+4}} =[/tex]

    from here the denominator looks similar to e and the numerator almost looks like e, however I’m a little hesitant to call the limit e/(e^4) = 1/(e^3) therefore I applied bernouli’s inequality to the numerator :

    [tex] \geq \lim_{n\rightarrow\infty}\frac{1 + (n + 4)(3/n)}{e^{4}} \geq \frac{\lim_{n\rightarrow\infty}(1 + 3 + (12/n))}{e^{4}} \geq \frac{4}{e^{4}}[/tex]

    Is my thinking correct in operating this way?
    Last edited: Oct 24, 2005
  6. Oct 25, 2005 #5


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    No. Why don't you just write it out and use some limit rules. This looks like the limit of a product of 2 factors, both with limits you know.

    Don't just guess what you think the limit should be. What makes you think that:
    [tex]\lim_{n\to \infty} \left(1+\frac{1}{n}\right)^{n+4}=e^4[/tex]
    ? You can solve this limit in the same way as the previous one by considering it a product of 2 limits.

    For a limit like: [itex]\lim_{n\to \infty} (1+x/n)^n[/itex] which looks alot like the one for e, what method would be useful? (Hint: substitution)

    Another way is to write:
    Last edited: Oct 25, 2005
  7. Oct 26, 2005 #6
    thank you , i have found my errors
    [tex]\lim_{n\rightarrow\infty}(\frac{n+1}{n})^{n+1} = \lim_{n\rightarrow\infty}(\frac{n}{n} + \frac{1}{n})^{n+1} = \lim_{n\rightarrow\infty}(1 + \frac{1}{n})^{n+1}\\[/tex]
    [tex]\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^n \left(1+ \frac{1}{n}\right) = e \times \lim_{n\rightarrow\infty}\left(1+ \frac{1}{n}\right) = e[/tex]
    for the other one:
    [tex]\lim_{n\rightarrow\infty}\left(\frac{n+3}{n+1}\right)^{n+4}= \lim_{n\to \infty} \left(1+\frac{2}{n+1}\right)^{n+4} = \lim_{n\rightarrow\infty}\left(1 + \frac{2}{n+1}\right)^{n+1}\left(1 + \frac{2}{n+1}\right)^{3} = \lim_{n\rightarrow\infty}\left(1 + \frac{2}{n+1}\right)^{n+1} \times 1[/tex]

    now i have found that this actually goes to [tex]e^{2}[/tex] but i did get some help but i don't know how to prove it other than writing out the sequence values for e and then for this sequence [tex]\lim_{n\rightarrow\infty}\left(1 + \frac{2}{n+1}\right)^{n+1} [/tex]
    and noticing that e is a subsequence of the above
    Last edited: Oct 26, 2005
  8. Oct 26, 2005 #7


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    You can make a substitution: m=n/x. That makes:

    [tex]\lim_{n\to \infty} \left(1+\frac{x}{n}\right)^n[/tex]
    look like:
    [tex]\lim_{m\to \infty}\left(1+\frac{1}{m}\right)^{xm}[/tex]
    Or m-> -infinity if x is negative.
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