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Math Help - Limit continuity

  1. #1
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    Limit continuity

    Hey guys. Would appreciate a little help woth the following problem. Im sure its quite straight forward but im not sure on how to approach it. Well here it is:

    lim <br />
         \frac{e^{2x}-2e^x+1}{x^2}<br />
    x->0

    Thanks a lot for any help.
    Last edited by ledbymischief; March 8th 2008 at 08:53 AM.
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  2. #2
    Super Member PaulRS's Avatar
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    Note that: (e^{2x}-2e^x+1)=(e^x-1)^2

    Thus: \frac{e^{2x}-2e^x+1}{x^2}=\left(\frac{e^x-1}{x}\right)^2

    Now remember that: \lim_{u\rightarrow{0}}\frac{e^u-1}{u}=1

    So, by the continuity of g(x)=x^2 we have that \lim_{x\rightarrow{0}}\frac{e^{2x}-2e^x+1}{x^2}=\lim_{x\rightarrow{0}}\left(\frac{e^x-1}{x}\right)^2=\left(\lim_{x\rightarrow{0}}\frac{e  ^x-1}{x}\right)^2=1
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  3. #3
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    Thanks Paul. I can't believe i missed that. Great help!
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