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Math Help - L'Hopital's Rule

  1. #1
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    L'Hopital's Rule

    Can anyone show a simple proof of L'Hopital's Rule?

    lim as x->0 of f(x)/g(x) = f'(0)/g'(0)

    Thanks...
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by jzellt View Post
    Can anyone show a simple proof of L'Hopital's Rule?

    lim as x->0 of f(x)/g(x) = f'(0)/g'(0)

    Thanks...
    The proof of the general case is a bit ugly (at least the one I know is). Here's a simple proof of the \frac{0}{0} case of the rule:

    Theorem: If f(x_0)=0=g(x_0), f'(x_0) and g'(x_0) exist and do not both equal 0 or \infty, then \lim_{x\to x_0}\frac{f(x)}{g(x)}=\frac{f'(x_0)}{g'(x_0)}.

    Proof: \lim_{x\to x_0}\frac{f(x)}{g(x)}=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{g(x)-g(x_0)}=\lim_{x\to x_0}\frac{\displaystyle\frac{f(x)-f(x_0)}{x-x_0}}{\displaystyle\frac{g(x)-g(x_0)}{x-x_0}}= \frac{\displaystyle\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}}{\displaystyle\lim_{x\to x_0}\frac{g(x)-g(x_0)}{x-x_0}}=\frac{f'(x_0)}{g'(x_0)}

    The \frac{\infty}{\infty} case is trickier.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    redsoxfan325's proof is quite a common reasoning, but for a rigorous proof you need a bit of machinery. Namely, what really is a derivative? etc.

    @redsoxfan325- I of course in no way am implying that you don't know that (I have seen your posts and you obviously do) but am merely making an observation for the OP.
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