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Math Help - Indeterminate Form

  1. #1
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    Indeterminate Form

    x -> 0

    What is the indeterminate form of

    lim n-> infinity (1 + 7/n + x/n)^n

    If I let u = (7+x)/n, the problem becomes

    lim n-> infinity (1+u)^((7+x)/u)

    And the answer is e^7, but I don't understand the process of obtaining the answer.
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  2. #2
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    Quote Originally Posted by soma View Post
    x -> 0

    What is the indeterminate form of

    lim n-> infinity (1 + 7/n + x/n)^n

    If I let u = (7+x)/n, the problem becomes

    lim n-> infinity (1+u)^((7+x)/u)

    And the answer is e^7, but I don't understand the process of obtaining the answer.
    Actually the answer is e^{7 + x}.

    You have \lim_{n \rightarrow + \infty} \left( 1 + \frac{7 + x}{n}\right)^n = \lim_{n \rightarrow + \infty} \left( 1 + \frac{a}{n}\right)^n = e^a.

    (The last one is a standard limit and should be known).
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  3. #3
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    for:

    \lim_{n \rightarrow + \infty} \left( 1 + \frac{a}{n}\right)^n = e^a

    How can I show the work on my homework assignment for that particular step? I know you say it's a standard limit, but I can't find any information about it online... Or is it just a matter of saying it's a standard limit?
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  4. #4
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    Quote Originally Posted by soma View Post
    for:

    \lim_{n \rightarrow + \infty} \left( 1 + \frac{a}{n}\right)^n = e^a

    How can I show the work on my homework assignment for that particular step? I know you say it's a standard limit, but I can't find any information about it online... Or is it just a matter of saying it's a standard limit?
    Here's a site

    http://en.wikipedia.org/wiki/E_(mathematical_constant)

    As for your problem, from

     \lim_{n \to + \infty} \left( 1 + \frac{1}{n}\right)^n = e

    set n = \frac{m}{a} noting that n \to \infty gives m \to \infty so

     \lim_{m \to + \infty} \left( 1 + \frac{a}{m}\right)^{m/a} = e

    so

     \left(\lim_{m \to + \infty} \left( 1 + \frac{a}{m}\right)^{m/a} \right)^a = e^a.
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  5. #5
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    That's perfect. I now understand the problem completely, thanks!
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