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Math Help - L'Hospital's Rule

  1. #1
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    L'Hospital's Rule

    Hi, i'm trying to help my sibling with these math problems, but i'm having some trouble:

    1. lim x-> -infinity (x^2)^(e^x)

    2. and lim x -> 0 (1/x-1/ln(x+1))



    For one 1. I use the natural log to make e^x * ln(x^2) and i've tried put both over eachother but it goes nowhere. e^x will continue to be zero. I've used the rule repeatedly and cannot find an answer. I don't see part of the derivative of ln(x^2) or 1/ln(x^2) helping yet either.

    For two i've put both over a common denominator but still am having trouble getting an answer

    thanks
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  2. #2
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    Hello,
    Quote Originally Posted by jarny View Post
    Hi, i'm trying to help my sibling with these math problems, but i'm having some trouble:

    1. lim x-> -infinity (x^2)^(e^x)

    2. and lim x -> 0 (1/x-1/ln(x+1))



    For one 1. I use the natural log to make e^x * ln(x^2) and i've tried put both over eachother but it goes nowhere. e^x will continue to be zero. I've used the rule repeatedly and cannot find an answer. I don't see part of the derivative of ln(x^2) or 1/ln(x^2) helping yet either.
    Hmmm what about this :

    e^x \ln(x^2)=\frac{\ln(x^2)}{\frac{1}{e^x}}=\frac{\ln(  x^2)}{e^{-x}}

    as x goes to - infinity, x^2 goes to + infinity and hence ln(x^2) goes to + infinity
    as x goes to - infinity, -x goes to + infinity and hence e^{-x} goes to + infinity.

    So now you can apply the rule. Just differentiate (by using chain rule) and see if it can help.

    For two i've put both over a common denominator but still am having trouble getting an answer
    I guess it's better you show what you've done =)
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  3. #3
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    Hello, jarny!

    I think I got #2 . . .


    2)\;\;\lim_{x\to0}\left[\frac{1}{x}-\frac{1}{\ln(x+1)}\right]
    Combine the fractions: . \frac{\ln(x+1)-x}{x\ln(x+1)}


    Since this goes to \frac{0}{0}. we can apply L'Hopital.

    . . Then we have: . \frac{\frac{1}{x+1} - 1}{x\cdot\frac{1}{x+1} + \ln(x+1)}


    Multiply by \frac{x+1}{x+1}\!:\quad \frac{1 - (x+1)}{x + (x+1)\ln(x+1)} \;=\;\frac{-x}{x+(x+1)\ln(x+1)}


    Since this goes to \frac{0}{0}, we can apply L'Hopital again.

    . . Then we have: . \frac{-1}{1 + (x+1)\cdot\frac{1}{x+1} + \ln(x+1)} \;=\;\frac{-1}{2 + \ln(x+1)}


    Therefore: . \lim_{x\to0}\left[\frac{-1}{2+\ln(x+1)}\right] \;=\;\frac{-1}{2+0} \;=\;\boxed{-\frac{1}{2}}

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