# How to prove $\lim_{h \to 0} θ ＝ \frac{1}{n＋1}$?

$f(x)$ has a continuous derivative of $n＋1$ order, and$$f(a＋h)＝f(a)＋hf'(a)＋\frac{h^2}{2!}f''(a)＋ \cdots ＋\frac{h^n}{n!} f^{(n)}(a＋θh). \quad 0＜θ＜1, f^{(n + 1)}(a)≠0.$$ I know that for this question the Taylor formula should be used, but I don't know how to solve for it.

We have $$f(a+h)-f(a)=\displaystyle\sum_{i=1}^{n-1} {f^{(i)}(a)\over i!}\cdot h^i+{f^{(n)} (a+\theta h)\over n!}\cdot h^n$$$$=\sum_{i=1}^{n} {f^{(i)}(a)\over i!}\cdot h^i+{f^{(n+1)}(a+\xi h)\over (n+1)!}\cdot h^{n+1} ,$$ Here $\theta,\xi\in (0,1).$ Thus $$f^{(n)} (a+\theta h)=f^{(n)}(a)+\displaystyle{f^{(n+1)}(a+\xi h)\over (n+1)}\cdot h.$$ Notice that $f^{(n)}(a+\theta h)-f^{(n)}(a)=f^{(n+1)}(a+\eta h)\cdot \theta h,$ here $\eta\in (0,\theta),$ thus we have $$f^{(n+1)}(a+\eta h)\cdot \theta=\displaystyle{f^{(n+1)}(a+\xi h)\over (n+1)}.$$ Now let $h\to 0,$ and we obtain that $\theta\to \displaystyle {1\over n+1}.$

• Does not touch the solution but shouldn't $i$ start from 1 rather than 0? – LRDPRDX Jan 15 '18 at 9:41
• @BogdanSikach: You're right, thanks. I'll edit my answer. – painday Jan 15 '18 at 9:52