Taylor's theorem Cauchy's remainder question. I was studying the rigorous definition of Taylor's theorem and various kinds of remainders' forms when I got stuck at this question.
If $f^{ii}(a)=f^{iii}(a)=f^{iv}(a)=.....=f^{n-1}(a)=0$ but $f^{n}(x)$ is continuous non zero at $x=a$, the we need to prove that $$  \lim_{h \to 0} (\theta _{n -1})=1/n$$ 
where $$f(a+h)=f(a)+hf'(a)+.......+\frac{h^{n-1}}{(n-1)!} f^{n-1} (x+\theta _{n-1}h)$$ 
I tried to use the Cauchy's remainder form by writing
$$R_n= \frac{h^{n-1}(1-\theta)^{n-1} }{(n-1)!} f^n(a+ \theta h)$$
For the sequence to converge now $$ \lim_{n \to \infty} R_n = 0$$ which seems to be perfectly true because of the $\frac{h^{n-1}}{(n-1)!}$ and I use limit of $h$ tending to zero, that just makes the whole expression $0$ at once. How can I reach the final answer?
P.S. Though my book mentions the limit in the question as $h$ tending to zero but I just can't fathom as to how can $\theta$ depend on $h's$ decreasing value.
 A: I don't like to insist, but it seems a missing brace. Anyway, it's true that,
$$\lim_{h\to0}\theta_{n-1}=\frac{1}{n}$$
Fristly, from the hypothesis with the remainder of order $n-1$:
$$f(a+h)=f(a)+hf'(a)+\dfrac{h^{n-1}}{(n-1)!} f^{(n-1)} (a+\theta _{n-1}h)$$
$\dfrac{h^{n-1}}{(n-1)!} f^{(n-1)} (a+\theta _{n-1}h)=f(a+h)-f(a)-hf'(a)\tag 1$
And isolating $\theta_{n+1}$ in the mean value theorem applied to the $n-1$ derivative:
$f^{(n-1)} (a+\theta _{n-1}h)-f^{(n-1)}(a)=\theta_{n-1}hf^{(n)}(\xi)$, with $\xi\in(a,a+\theta_{n-1}h)$. So is,
$f^{(n-1)} (a+\theta _{n-1}h)=\theta_{n-1}hf^{(n)}(\xi)$
$\theta_{n-1}=\dfrac{f^{(n-1)} (a+\theta _{n-1}h)}{hf^{(n)}(\xi)}=\dfrac{\dfrac{h^{n-1}}{(n-1)!}f^{(n-1)} (a+\theta _{n-1}h)}{\dfrac{h^{n}}{(n-1)!}f^{(n)}(\xi)}$
Taking the limit:
$$\lim_{h\to0}\theta_{n-1}=(n-1)!\lim_{h\to0}\dfrac{\dfrac{h^{n-1}}{(n-1)!}f^{(n-1)} (a+\theta _{n-1}h)}{h^{n}f^{(n)}(\xi)}$$
But $f^{(n)}$ is continuous and $f^{(n)}(a)\ne0$, then $\lim_{h\to0}f^{(n)}(\xi)=f^{(n)}(a)$ and can be taken out from the limit operation.
$$\lim_{h\to0}\theta_{n-1}=\dfrac{(n-1)!}{f^{(n)}(a)}\lim_{h\to0}\dfrac{\dfrac{h^{n-1}}{(n-1)!}f^{(n-1)} (a+\theta _{n-1}h)}{h^{n}}=$$
$$\overset{(1)}=\dfrac{(n-1)!}{f^{(n)}(a)}\lim_{h\to0}\dfrac{f(a+h)-f(a)-hf'(a)}{h^{n}}$$
By using the L'Hôpital rule $n-1$ times, and remembering that $f^{(n-1)}(a)=0$:
$$\lim_{h\to0}\theta_{n-1}=\dfrac{(n-1)!}{f^{(n)}(a)}\lim_{h\to0}\dfrac{f^{(n-1)}(a+h)-f^{(n-1)}(a)}{n!h}=$$
$$=\dfrac{1}{nf^{(n)}(a)}\lim_{h\to0}\dfrac{f^{(n-1)}(a+h)-f^{(n-1)}(a)}{h}=$$
$$=\dfrac{1}{nf^{(n)}(a)}f^{(n)}(a)=\dfrac{1}{n}$$
