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Math Help - Two limit problems

  1. #1
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    Two limit problems

    My last pretty straight forward, How about these two.

    lim (1 +3h)^1/h
    h->0
    or

    lim 3^-x
    x->+infin
    Last edited by Jhevon; April 8th 2009 at 03:52 PM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Meeklo Braca View Post
    That one is pretty straight forward, How about these two.

    lim (1 +3h)^1/h
    h->0
    Note that \lim_{h \to 0} (1 + 3h)^{1/h} = \lim_{h \to 0} e^{\ln (1 + 3h)^{1/h}} = \lim_{h \to 0}e^{\frac {\ln (1 + 3h)}h} Now, apply L'Hopital's rule.


    lim 3^-x
    x->+infin
    recall that 3^{-x} = \frac 1{3^x}
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  3. #3
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    I have no idea how to apply L'Hopital's rule to the first question.

    With that question I did y = 1/h as h->0, y->0 therefore the lim (1+3h)^0 is 0

    As for the 2nd one, is that implying that the answer is 0?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Meeklo Braca View Post
    I have no idea how to apply L'Hopital's rule to the first question.
    do you know what L'Hopital's rule says? remember, \lim_{x \to a} e^{f(x)} = e^{\lim_{x \to a}f(x)}

    With that question I did y = 1/h as h->0, y->0 therefore the lim (1+3h)^0 is 0
    first of all, 1^0 is 1, not zero. secondly, y would tend to infinity as h tends to 0. thirdly, if you are changing h, you would have to change ALL the h's, meaning your new limit would be \lim_{y \to \infty} \left( 1 + \frac 3y \right)^y

    As for the 2nd one, is that implying that the answer is 0?
    um, yeah. we have a constant numerator and the denominator is growing without bound.
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