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Math Help - another limit of a sequence

  1. #1
    Member disclaimer's Avatar
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    another limit of a sequence

    Hello;

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

    Sorry if it's too simple, but how to calculate the limit above? All I can see right now is \left[\infty\cdot0\right].

    Thanks.
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  2. #2
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    Quote Originally Posted by disclaimer View Post
    Hello;

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

    Sorry if it's too simple, but how to calculate the limit above? All I can see right now is \left[\infty\cdot0\right].

    Thanks.
    Here.
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  3. #3
    Eater of Worlds
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    Here's one way.

    Let h=\frac{1}{n}, \;\ n=\frac{1}{h}

    Make the subs and change the limit to h-->0:

    \lim_{h\rightarrow{0}}\frac{e^{h}-1}{h}

    This is a famous limit.

    To show it we can express the derivative of e^x at x=0 as a limit.

    \frac{d}{dx}[e^{x}]|_{x=0}=\lim_{u\rightarrow{0}}\frac{e^{0+h}-e^{0}}{h}=\lim_{h\rightarrow{0}}\frac{e^{h}-1}{h}=1
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