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Math Help - Limits

  1. #1
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    Limits

    I am having trouble with this limit, and I would appreciate anyone's imput.

    limit (1+(1/3X))^(7+2X)
    as x->0

    I have tried to manipulate this into the 2 indeterminate forms that work with L'hopitals rule to no avail.
    I am leaning toward the limit approaching 0 , since the limit is undefined at x=0, if we approach x from 0+, the limit goes to infinity^7 , which means 1/[(inifinity)^-7)] = 0

    Thank you for your help
    Last edited by ewqewq; February 27th 2011 at 09:00 PM.
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  2. #2
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    Is this \displaystyle \lim_{x \to 0}\left[1 + \left(\frac{1}{3x}\right)^{7 + 2x}\right] or \displaystyle \lim_{x \to 0}\left[\left(1 + \frac{1}{3x}\right)^{7+2x}\right], or some other form I haven't written?
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  3. #3
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    It is the second one . I am sorry mine was not clear.
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  4. #4
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    \displaystyle \lim_{x \to 0}\left[\left(1 + \frac{1}{3x}\right)^{7+2x}\right] =  \lim_{x \to 0}\left\{e^{\ln{\left[\left(1+\frac{1}{3x}\right)^{7+2x}\right]}}\right\}

    \displaystyle = \lim_{x \to 0}\left[e^{(7+2x)\ln{\left(1 + \frac{1}{3x}\right)}}\right]

    \displaystyle = e^{\lim_{x \to 0}\left[(7+2x)\ln{\left(1 + \frac{1}{3x}\right)}\right]}.

    It doesn't look like you need to use L'Hospital's Rule at all...
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  5. #5
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    Thanks for your help, however I still do not understand what would happen to 1/(3x), since it would be undefined
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