Tom Apostol mathematical analysis exercise 5.28 Prove the following theorem do not use L‘Hospital rule 
5.28
  Let $f$ and $g$ be two functions having finite nth derivatives in (a,b).for some interior point c in (a,b),assumes that $f(c)=f’(c)=\dots=f^{n-1}(c)=0$,and that$g(c)=g’(c)=\dots=g^{n-1}(c)=0$,but $g^{n}(x)$is never zero in (a,b),show that
  $$\lim_{x\to c}\frac{f(x)}{g(x)}=\frac{f^{n}(c)}{g^{n}(c)}$$
Note. $f^{(n)}$ and $g^{(n)}$ are  not assumed to be continous at c. $Hint.$ Let
$$F(x)=f(x)-\frac{(x-c)^{n-1}f^{(n-1)}(c)}{(n-1)!},$$

I research for n=1.
$F(x)=f(x)-f(c),G(x)=g(x)-g(c)$,use Cauchy mean value theorem. I got
$f(x)g’(x_1)=f’(x_1)g(x)$ where $x_1$ is interior to the interval joining $x$ and $c$
$$|\frac{f(x)}{g(x)}-\frac{f’(c)}{g’(c)}|=|\frac{f’(x_1)}{g’(x_1)}-\frac{f’(c)}{g’(c)}|$$
But $f’(x),g’(x)$ are not assumed to be continuous at $c$. I don’t know how to continue.Could you help me ?pleasa do not use L Hospital because this exercise in which chapter isn’t teach L Hospital.
Thanks in advance!!!
PS,The problem in the book where emphasis that $f^{n}(x),g^{n}(x)$ is not assumed to be continuous at point c
 A: The expressions $f^{(k)}(c)$ and  $g^{(k)}(c)$ for $k \ge 1$ only make sense if $f^{(k-1)}$ and $g^{(k-1)}$ exist in a neighborhood of $c$. And you say that $g^{(n)}(x)$ is never $0$ in $(a,b)$ which means that $g$ must be $n$-times differentiable on $(a,b)$. So we may assume w.l.o.g. that both $f, g$ are $n$-times differentiable on $(a,b)$.
The mean value theorem implies that for $x \in J = (a,b) \setminus \{c\}$
$$g^{(n-1)}(x) = g^{(n-1)}(x) - g^{(n-1)}(c) = g^{(n)}(\xi)(x-c)$$
for some $\xi$ between $x$ and $c$. Since $g^{(n)}(\xi) \ne 0$ on $J \subset (a,b)$, this shows that $g^{(n-1)}(x) \ne 0$ for $x \in J$. Proceeding inductively shows that also $g^{(n-2)},\dots, g^{(1)}= g', g^{(0)} = g$ are nonzero on $J$. Hence the quotients
$$\frac{f^{(k)}(x)}{g^{(k)}(x)}$$
are defined on $J$ for $k =0,\dots, n$.
Let us first treat the case $n = 1$. We have $f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x-c} = \lim_{x \to c} \frac{f(x)}{x-c}$ and  $g'(c) = \lim_{x \to c} \frac{g(x)}{x-c}$, hence
$$\frac{f'(c)}{g'(c)} = \dfrac{\lim_{x \to c} \frac{f(x)}{x-c}}{\lim_{x \to c} \frac{g(x)}{x-c}} = \lim_{x \to c} \dfrac{ \frac{f(x)}{x-c}}{\frac{g(x)}{x-c}} = \lim_{x \to c} \dfrac{f(x)}{g(x)} .$$
Applying this to $f^{(n-1)}$ and $g^{(n-1)}$ yields
$$\frac{f^{(n)}(c)}{f^{(n)}(c)} =  \lim_{x \to c} \dfrac{f^{(n-1)}(x)}{g^{(n-1)}(x)} .$$
The extended mean value theorem (Cauchy's mean value theorem) says that for each $x \in J$ there exists $\xi$ between $x$ and $c$ such that
$$f^{(n-2)}(x)g^{(n-1)}(\xi) = (f^{(n-2)}(x)-f^{(n-2)}(c))g^{(n-1)}(\xi) = (g^{(n-2)}(x)-g^{(n-2)}(c))f^{(n-1)}(\xi) = g^{(n-2)}(x)f^{(n-1)}(\xi) .$$
Therefore
$$\frac{f^{(n-2)}(x)}{g^{(n-2)}(x)} = \frac{f^{(n-1)}(\xi)}{g^{(n-1)}(\xi)} .$$
If $x \to c$, then also $\xi \to c$ and we conclude
$$\lim_{x \to c} \frac{f^{(n-2)}(x)}{g^{(n-2)}(x)} = lim_{\xi \to c}\frac{f^{(n-1)}(\xi)}{g^{(n-1)}(\xi)} = \frac{f^{(n)}(c)}{g^{(n)}(c)} .$$
Proceeding inductively we get
$$\lim_{x \to c} \frac{f(x)}{g(x)} = \frac{f^{(n)}(c)}{g^{(n)}(c)} .$$
