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Math Help - [SOLVED] Limit when Hopital's rule fails

  1. #1
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    [SOLVED] Limit when Hopital's rule fails

    I have a function f(x) = g^{-1}(h(x)) and I'm looking for \lim_{x \rightarrow x_0^+} f'(x)

    I know the following about the functions g and h:
    • h(x_0)=0 and g(0) = 0 so f(x_0)=0
    • g \geq 0, h \geq 0
    • g'(0)=0, h'(x_0)=0
    • g' > 0 for x \neq x_0, h' > 0 for x<x_0 and h'<0 for x>x_0
    • g'' > 0, h'' > 0 and finite at x_0
    • h''' = 0, g''' finite at x_0
    We have f'(x)= \frac{h'(x)}{g'(f(x))}. At x_0, both numerator and denominator are 0. Applying l'Hopital's rule, we obtain

    \lim_{x \rightarrow x_0^+} f'(x) = \lim_{x \rightarrow x_0^+} \frac{h''(x)}{f'(x) g''(f(x))} = \lim_{x \rightarrow x_0^+} \frac{h''(x) g'(f(x))}{h'(x) g''(f(x))} .

    Applying l'Hopital's rule again, we're back to where we started:

     \lim_{x \rightarrow x_0^+} \frac{h'''(x) g'(f(x))+h''(x) f'(x) g''(f(x))}{h''(x) g''(f(x)) + h'(x) f'(x) g'''(f(x))} = \lim_{x \rightarrow x_0^+} f'(x).

    There's probably a simpler way of presenting this problem, sorry about that... If anyone knows how this kind of circularity problem is called and/or how to deal with it, I will be very grateful!
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  2. #2
    Moo
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    Hello,

    <br />
\lim_{x \rightarrow x_0^+} f'(x) = \lim_{x \rightarrow x_0^+} \frac{h''(x)}{f'(x) g''(f(x))}<br />
    When you're here, why don't you multiply both sides by f'(x) ? (if I'm correct, it's not important whether f'(x)=0 or not).

    \lim_{x \to x_0^+} (f'(x))^2=\lim_{x \to x_0^+} \frac{h''(x)}{g''(f(x))}

    h'' and g'' are different from 0, positive and finite at x_0, according to what you've written.

    So :

    \lim_{x \to x_0^+} f'(x)=\sqrt{\frac{h''(x_0)}{g''(f(x_0))}}

    But we know that f(x_0)=0.

    Thus \lim_{x \to x_0^+} f'(x)=\sqrt{\frac{h''(x_0)}{g''(0)}}
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    Quote Originally Posted by Moo View Post

    When you're here, why don't you multiply both sides by f'(x) ? (if I'm correct, it's not important whether f'(x)=0 or not).

    \lim_{x \to x_0^+} (f'(x))^2=\lim_{x \to x_0^+} \frac{h''(x)}{g''(f(x))}
    I had tried this method with a couple of examples in Mathematica, and it didn't seem to give me correct results. Are you sure that it's correct to do that manipulation?
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    Moo
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    Quote Originally Posted by flowtsen View Post
    I had tried this method with a couple of examples in Mathematica, and it didn't seem to give me correct results. Are you sure that it's correct to do that manipulation?
    My fault, it is important whether f '(x_0)=0 or not.

    If it is not, use the method above, if it is, the limit is obviously 0

    Do you have examples ?
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    After double-checking, this method indeed seems to work. Thanks very much, especially for the quickness of your response!
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