Differentiation

• Jan 12th 2011, 07:44 AM
Markeur
Differentiation
Let $\displaystyle I \subseteq \mathbb{R}$ be an interval and let $\displaystyle c \in I$. Suppose that $\displaystyle f$ and $\displaystyle g$ are defined on $\displaystyle I$ and that the derivatives $\displaystyle f^{(n)}$,$\displaystyle g^{(n)}$ exist and are continuous on $\displaystyle I$. If $\displaystyle f^{(k)}(c)=0$ and $\displaystyle g^{(k)}(c)=0$ for $\displaystyle k=0,1,...,n-1$, but $\displaystyle g^{(n)}(c) \neq 0$, show that
$\displaystyle \lim_{x \to c} \frac{f(x)}{g(x)} = \frac{f^{(n)}(c)}{g^{(n)}(c)}$
• Jan 12th 2011, 08:16 AM
roninpro
Maybe you could try Taylor's theorem? Write

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

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

for some $\displaystyle \xi_f \in (x,c)$ and $\displaystyle \xi_g \in (x,c)$. Taking the limit of the quotient should give you the result.
• Jan 13th 2011, 10:55 PM
Drexel28
Quote:

Originally Posted by Markeur
Let $\displaystyle I \subseteq \mathbb{R}$ be an interval and let $\displaystyle c \in I$. Suppose that $\displaystyle f$ and $\displaystyle g$ are defined on $\displaystyle I$ and that the derivatives $\displaystyle f^{(n)}$,$\displaystyle g^{(n)}$ exist and are continuous on $\displaystyle I$. If $\displaystyle f^{(k)}(c)=0$ and $\displaystyle g^{(k)}(c)=0$ for $\displaystyle k=0,1,...,n-1$, but $\displaystyle g^{(n)}(c) \neq 0$, show that
$\displaystyle \lim_{x \to c} \frac{f(x)}{g(x)} = \frac{f^{(n)}(c)}{g^{(n)}(c)}$

This is just a repeated (legitimate) application of L'hopital's rule.