1. ## 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
$\lim_{x \to c} \frac{f(x)}{g(x)} = \frac{f^{(n)}(c)}{g^{(n)}(c)}$

2. 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.

3. Originally Posted by Markeur
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.