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Math Help - Lemma that is used for L'hospital's rule

  1. #1
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    Lemma that is used for L'hospital's rule

    This proof is left as an exercise in the book:

    Suppose that g is continuously differentiable in an interval about x_0 and g(x_0) = 0. If g'(x_0) \neq 0, show that there is a small interval about x_0 in which neither g nor g' vanish except for the root of g at the point x_0.

    I don't really know what this is saying, nor how to prove to prove it.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by paulrb View Post
    This proof is left as an exercise in the book:

    Suppose that g is continuously differentiable in an interval about x_0 and g(x_0) = 0. If g'(x_0) \neq 0, show that there is a small interval about x_0 in which neither g nor g' vanish except for the root of g at the point x_0.

    I don't really know what this is saying, nor how to prove to prove it.
    What don't you understand about the statement? It's saying suppose that you found some point x_0 such that g(x_0)=0 yet g'(x_0)\ne 0 can you find some neighborhood of x_0 such that g' and g are neither zero anywhere on that neighborhood except g is of course zero at x_0. Anyways..

    Since g'(x_0)\ne 0 and g' is continuous there exists an open ball B_\delta(x_0) such that g' does not vanish on that open ball. Now, if g vanished somewhere else on that open ball besides x_0 you could apply Rolle's theorem to find a zero of g' on the open ball contradictory to construction.
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    Thanks...although we have not covered what an open ball is in the class. We have covered Rolle's theorem though.

    Intuitively I think I understand now. Since g'(x_0) \neq 0, there has to be some point near x_0where g is not 0. Since g is continuous, all of the points between x_0 and this point are nonzero. Furthermore for g to reach this nonzero point from 0, there has to be an interval between them where g'(x) is nonzero, since g'(x) is continuous. Although I still don't know how to prove this rigorously.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by paulrb View Post
    Thanks...although we have not covered what an open ball is in the class. We have covered Rolle's theorem though.

    Intuitively I think I understand now. Since g'(x_0) \neq 0, there has to be some point near x_0where g is not 0. Since g is continuous, all of the points between x_0 and this point are nonzero. Furthermore for g to reach this nonzero point from 0, there has to be an interval between them where g'(x) is nonzero, since g'(x) is continuous. Although I still don't know how to prove this rigorously.
    Hmm, I think you're mixing up the argument a little bit. Let me rephrase my argument in nicer language. We know that g'(x_0)\ne 0, right? Let's assume just for ease that g'(x_0)>0. Then, by continuity of g' there exists some \delta>0 such that \displaystyle |x-x_0|<\delta\implies \left|g(x_0)-g(x)\right|<\frac{g(x_0)}{2} which in particular implies that if |x-x_0|<\delta then g'(x)\ne 0 (check this). Thus, we have on the interval (x_0-\delta,x_0+\delta) that g' is not zero. Now, I claim that except for x_0 g(x)\ne 0 for x\in(x_0-\delta,x_0+\delta). Why? If x<x_0\in (x_0-\delta,x_0+\delta) such that g(x)=0 then there exists some \xi\in(x,x_0)\subseteq(x_0-\delta,x_0+\delta) such that g'(\xi)=0 but this contradicts that g is never zero on (x_0-\delta,x_0+\delta) and similarly g(x)\ne 0 for any x>x_0\in(x_0-\delta,x_0+\delta). Make sense?
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    Oh I get it now, thank you very much!
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