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Math Help - Finding the Limit

  1. #1
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    Finding the Limit

    Given lim S_n=(1+\dfrac{1}{n})^n \rightarrow e

    Find
    1)lim S_n=(1+\dfrac{1}{2n})^{2n} \rightarrow ?

    2)lim S_n=(1+\dfrac{1}{n})^{2n} \rightarrow ?

    Attempt:
    I see that \dfrac{1}{2n} converges to zero which makes it no different than \dfrac{1}{n} so we're left with S_n=(1+\dfrac{1}{n})^{2n} \rightarrow ? which should make both problems #1 and #2 have a limit of e^2.

    However, the back of the book indicates that #1 has a limit of e and #2 has a limit of e^2

    Help?
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  2. #2
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    Re: Finding the Limit

    Quote Originally Posted by spruancejr View Post
    Given lim S_n=(1+\dfrac{1}{n})^n \rightarrow e
    Find
    1)lim S_n=(1+\dfrac{1}{2n})^{2n} \rightarrow ?

    2)lim S_n=(1+\dfrac{1}{n})^{2n} \rightarrow ?

    Attempt:
    I see that \dfrac{1}{2n} converges to zero which makes it no different than \dfrac{1}{n} so we're left with S_n=(1+\dfrac{1}{n})^{2n} \rightarrow ? which should make both problems #1 and #2 have a limit of e^2.

    However, the back of the book indicates that #1 has a limit of e and #2 has a limit of e^2
    The text is correct in both cases.
    Here is a simple rule: {\left( {1 + \frac{a}{{bn + c}}} \right)^{dn}} \to {e^{\frac{{ad}}{b}}}
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  3. #3
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    Re: Finding the Limit

    Thanks!
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  4. #4
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    Re: Finding the Limit

    Quote Originally Posted by spruancejr View Post
    Given lim S_n=(1+\dfrac{1}{n})^n \rightarrow e

    Find
    1)lim S_n=(1+\dfrac{1}{2n})^{2n} \rightarrow ?

    2)lim S_n=(1+\dfrac{1}{n})^{2n} \rightarrow ?

    Attempt:
    I see that \dfrac{1}{2n} converges to zero which makes it no different than \dfrac{1}{n} so we're left with S_n=(1+\dfrac{1}{n})^{2n} \rightarrow ? which should make both problems #1 and #2 have a limit of e^2.

    However, the back of the book indicates that #1 has a limit of e and #2 has a limit of e^2

    Help?
    Your book is correct. For 1) if you let \displaystyle \begin{align*} t = 2n \end{align*} and note that as \displaystyle \begin{align*} n \to \infty, t \to \infty \end{align*}, then \displaystyle \begin{align*} \lim_{n \to \infty}\left(1 + \frac{1}{2n}\right)^{2n} = \lim_{t \to \infty}\left(1 + \frac{1}{t}\right)^t \end{align*}, which is the exact same limit as you were given in the first place.

    The reason 2) is \displaystyle \begin{align*} e^2 \end{align*} is because \displaystyle \begin{align*} \left(1 + \frac{1}{n}\right)^{2n} = \left[\left(1 + \frac{1}{n}\right)^n\right]^2 \end{align*}.
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