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Math Help - l'Hopital's rule

  1. #1
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    l'Hopital's rule

    Find <br />
    \lim_{x \to 0^{+}}\frac{x \log{x}}{\log{(1+2x)}}
    Can I do this:
    <br />
\lim_{x \to 0^{+}}\frac{x \log{x}}{\log{(1+2x)}}=<br />
\frac{\lim_{x \to 0^{+}} x \log{x}}{\lim_{x \to 0^{+}} \log{(1+2x)}}=\frac{\lim_{x \to 0^{+}} \frac{\log{x}}{\frac{1}{x}}}{\lim_{x \to 0^{+}} \log{(1+2x)}}=\frac{\lim_{x \to 0^{+}} -x}{\lim_{x \to 0^{+}} \log{(1+2x)}}<br />
    where l'Hopital's rule is applied to the numerator only. Then reapply l'Hopital's rule again:
    <br />
\lim_{x \to 0^+} \frac{-x}{\log (1+2x)}=\lim_{x \to 0^+} \frac {-1}{\frac{2}{1+2x}}=-\frac{1}{2}<br />

    Lemme know if my divide and conquer approach with l'Hopital's rule is legit.
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  2. #2
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    \displaystyle \lim_{x \to 0^+} \frac{x}{\log(1+2x)} \cdot \lim_{x \to 0^+} \log{x}

    \displaystyle\lim_{x \to 0^+} \frac{1+2x}{2} \cdot \lim_{x \to 0^+} \log{x}

    \displaystyle \frac{1}{2} \cdot \lim_{x \to 0^+} \log{x}

    the limit DNE
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  3. #3
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    so divide and conquer with l'Hopital and reapplying it is not legal?
    Last edited by crossbone; February 22nd 2011 at 05:49 PM.
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  4. #4
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    Quote Originally Posted by crossbone View Post
    so divide and conquer with l'Hopital and reapplying it is not legal?
    I would say not ... the original limit does not exist.

    limit as x to 0 &#40;x&#42;log&#40;x&#41;&#41;&#47;&#40;log&#40;1& #43;2x&#41;&#41; - Wolfram|Alpha
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  5. #5
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    Here's another problem: \displaystyle \lim_{x \to 0} x |{\log x|^a
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  6. #6
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    Rewrite it as \displaystyle \lim_{x \to 0}\frac{|\log{x}|^a}{\frac{1}{x}}, which is now of the form \displaystyle \frac{\infty}{\infty} so you can now use L'Hospital's Rule...
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  7. #7
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    I knew that but this is what I got:
    <br />
\displaystyle \lim_{x \to 0}\frac{|\log{x}|^a}{\frac{1}{x}}=\lim_{x \to 0} \frac{a|\log{x}|^{a-1}}{\frac{-1}{x}}<br /> <br />
    keep differentiating until a-n=b<0:
    <br />
\displaystyle \lim_{x \to 0} {(-)}^n k \frac{|\log{x}|^b}{\frac{1}{x}}=\lim_{x \to 0} {(-)}^n k x|\log{x}|^b=0<br />
    where
    <br />
k=a\cdot(a-1)\cdot(a-2)\cdot (a-3).....(a-n+1)<br />

    Is this right?
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  8. #8
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    The derivative of \displaystyle |\log{x}|^a is not \displaystyle a|\log{x}|^{a-1}.

    You need to use the chain rule...
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  9. #9
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by skeeter View Post
    What exactly does this approach violate? I'm looking at the rules for L`Hopitals and I don't quite see where this goes wrong, but it obviously does violate something: http://bradley.bradley.edu/~delgado/121/Lhopital.pdf

    Any ideas?
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  10. #10
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by crossbone View Post
    Find <br />
    \lim_{x \to 0^{+}}\frac{x \log{x}}{\log{(1+2x)}}
    Can I do this:
    <br />
\lim_{x \to 0^{+}}\frac{x \log{x}}{\log{(1+2x)}}=<br />
\frac{\lim_{x \to 0^{+}} x \log{x}}{\lim_{x \to 0^{+}} \log{(1+2x)}}=\frac{\lim_{x \to 0^{+}} \frac{\log{x}}{\frac{1}{x}}}{\lim_{x \to 0^{+}} \log{(1+2x)}}=\frac{\lim_{x \to 0^{+}} -x}{\lim_{x \to 0^{+}} \log{(1+2x)}}<br />
    where l'Hopital's rule is applied to the numerator only.

    Right. With this, you prove


    \displaystyle \lim_{x \to 0^{+}}\dfrac{f(x)}{g(x)}=\ldots=\dfrac{0}{0}


    Then reapply l'Hopital's rule again:
    <br />
\lim_{x \to 0^+} \frac{-x}{\log (1+2x)}=\lim_{x \to 0^+} \frac {-1}{\frac{2}{1+2x}}=-\frac{1}{2}<br />

    Lemme know if my divide and conquer approach with l'Hopital's rule is legit.

    But now, you find:


    \displaystyle \lim_{x \to 0^{+}}\dfrac{f''(x)}{g'(x)}


    Fernando Revilla
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  11. #11
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    Quote Originally Posted by Prove It View Post
    The derivative of \displaystyle |\log{x}|^a is not \displaystyle a|\log{x}|^{a-1}.

    You need to use the chain rule...
    I did. it appears I didn't cos I simplified it after I used l'Hopital's rule:

    <br />
\displaystyle \lim_{x \to 0}\frac{|\log{x}|^a}{\frac{1}{x}}=\lim_{x \to 0} \frac{\frac{a}{x}|\log{x}|^{a-1}}{\frac{-1}{x^2}}=\lim_{x \to 0}\frac{a|\log{x}|^{a-1}}{\frac{-1}{x}}<br /> <br />
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