Problem: Let $f(x)$ be $q$ continuous and differentiable with $f(r)=f^{(1)}(r)=...=f^{(q-1)}(r)=0$ and $f^{(q)}(r)\neq 0.$ Show that $$u(x)=\frac{f(x)}{f'(x)}$$ fulfills $$u'(r)=\displaystyle\lim_{x\rightarrow r} \frac{u(x)}{x-r}=\frac{1}{q}\neq 0.$$
Attempt: I have tried to use Taylor series on $f$ and $f'$ such that: $$f(x)=f(r)+f'(r)(x-r)+...+\frac{f^{(q)}(r)}{q!}(x-r)^q$$ and $$f'(x)=f'(r)+f''(r)(x-r)+...+q\cdot \frac{f^{(q-1)}(r)}{(q-1)!}(x-r)^{q-1}$$
But when i put in the formula given i keep end up dividing by zero, because $f^{(q-1)}(r)=0.$
Another attempt is that i'm trying to use MVT, but i then again run into the problem that i don't know anything about $f'(\xi).$
Any hints out there?
The condition you mention means that $r$ is a root with multiplicity $q$ and so you can use the factorisation $f(x)= (x-r)^{q} \phi(x)$, where $\phi(x) \ne 0$ in a neighbourhood of $r$.
$$ \lim_{x\to r}\frac{u(x)}{x-r} = \lim_{x\to r}\frac{(x-r)^q \phi(x)}{(x-r)(q(x-r)^{q-1} \phi (x) + (x-r)^q)\phi'(x)}=\lim_{x\to r}\dfrac{\phi(x)}{q \phi(x) + (x-r)\phi'(x)} = \frac{1}{q}. $$