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Math Help - Differentiation

  1. #1
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    Differentiation

    Let I \subseteq \mathbb{R} be an interval and let c \in I. Suppose that f and g are defined on I and that the derivatives f^{(n)}, g^{(n)} exist and are continuous on I. If f^{(k)}(c)=0 and g^{(k)}(c)=0 for k=0,1,...,n-1, but g^{(n)}(c) \neq 0, show that
    \lim_{x \to c} \frac{f(x)}{g(x)} = \frac{f^{(n)}(c)}{g^{(n)}(c)}
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  2. #2
    Senior Member roninpro's Avatar
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    Maybe you could try Taylor's theorem? Write

    \displaystyle f(x)=f(c)+f'(c)(x-c)+\frac{f''(c)}{2!}(x-c)^2+\ldots+\frac{f^{(n-1)}(c)}{(n-1)!}(x-c)^{n-1}+\frac{f^{(n)}(\xi_f)}{n!}(x-c)^n=\frac{f^{(n)}(\xi_f)}{n!}(x-c)^n

    \displaystyle g(x)=\frac{g^{(n)}(\xi_g)}{n!}(x-c)^n

    for some \xi_f \in (x,c) and \xi_g \in (x,c). Taking the limit of the quotient should give you the result.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Markeur View Post
    Let I \subseteq \mathbb{R} be an interval and let c \in I. Suppose that f and g are defined on I and that the derivatives f^{(n)}, g^{(n)} exist and are continuous on I. If f^{(k)}(c)=0 and g^{(k)}(c)=0 for k=0,1,...,n-1, but g^{(n)}(c) \neq 0, show that
    \lim_{x \to c} \frac{f(x)}{g(x)} = \frac{f^{(n)}(c)}{g^{(n)}(c)}
    This is just a repeated (legitimate) application of L'hopital's rule.
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