let $f(x)=f(x_{0})+f'(x_0)(x-x_0)+\cdots+\frac{f^{(n)}(\xi(x))}{n!}(x-x_{0})^{n}$
I need to prove that $x \mapsto f^{(n)}(\xi(x))$ is continuous if $f\in C^n[x_{0}-\delta,x_{0}+\delta]$ or in other words if $f$ is $n$ times differentiable.
A function is continuous if $\lim_{x\rightarrow x_{0}}f(x)=f(x_0)$. I know that $f$ must be continuous, because it is n times differentiable. However when we reveal $f^{(n)}(\xi(x))$ we get that
$f^{(n)}(\xi(x))=\frac{(f(x)-f(x_{0})-f'(x_0)(x-x_{0})-\frac{f''(x_0)(x-x_{0})^{2}}{2}-\cdots-f^{(n-1)}(x_0)\frac{(x-x_0)^{n-1}}{(n-1)!})n!}{(x-x_{0})^{n}}$
When we take this to the limit we get
$\lim_{x\rightarrow x_{0}}\frac{(f(x)-f(x_{0})-f'(x_0)(x-x_{0})-\frac{f''(x_0)(x-x_{0})^{2}}{2}-\cdots-f^{(n-1)}(x_0)\frac{(x-x_0)^{n-1}}{(n-1)!})n!}{(x-x_{0})^{n}}$ and from there a $\frac{0}{0}$ uncertainty.
This is the extent to which I think I have correctly solved this exercise, but now I (this is how I must solve it) must apply the L'Hospital rule. I am not sure what is the correct way to go about this, but I'll give what I think might be true.
If we differentiate both the numerator and the denominator once we get:
$\lim_{x\rightarrow x_{0}}\frac{(f'(x)-0-f'(x_0)-f''(x_0)(x-x_0)-\cdots-f^{(n-1)}(x_0)\frac{(x-x_0)^{n-2}}{(n-2)!})n!}{n(x-x_0)^{n-1}}$
Assuming this is even correct, I have no idea where to go from here. If at all possible, please provide explanations. Thanks!
