这个问题还是比较基础的一个taylor 展开问题，应该是书上的习题吧。
首先，题目提示你 \[{f^{\left( {n + 1} \right)}}\left( x \right) \ne 0\] ，所以你哟考虑 n+1阶展开
\[f\left( {x + h} \right) = f(x) + hf'(x) + .... + \frac{{{f^{n + 1}}(x + {\theta _2}h){h^{n + 1}}}}{{(n + 1)!}},\left( {0 < {\theta _2} < 1} \right)\]
又因为 n阶展开式为：
\[f\left( {x + h} \right) = f(x) + hf'(x) + .... + \frac{{{f^n}(x + {\theta _1}h){h^n}}}{{n!}},\left( {0 < {\theta _1} < 1} \right)\]
所以，有：
\[\frac{{{f^{\left( {n + 1} \right)}}(x + {\theta _2}h){h^{n + 1}}}}{{(n + 1)!}} + \frac{{{f^n}(x){h^n}}}{{n!}} = \frac{{{f^n}(x + {\theta _1}h){h^n}}}{{n!}}\]
所以：
\[\begin{array}{l} \mathop {\lim }\limits_{h \to 0} \frac{{{f^{\left( {n + 1} \right)}}(x + {\theta _2}h)}}{{n + 1}} = \mathop {\lim }\limits_{h \to 0} \frac{{{f^n}(x + {\theta _1}h) - {f^n}(x)}}{{{\theta _1}h}}{\theta _1}\\ {\rm{ \qquad\qquad \qquad }} = \mathop {\lim }\limits_{h \to 0} {f^{n + 1}}(x + {\theta _3}h){\theta _1},\qquad\left( {0 < {\theta _3} < 1} \right) \end{array}\]
所以：
\[{\theta _1} = \mathop {\lim }\limits_{h \to 0} \frac{{{f^{\left( {n + 1} \right)}}(x + {\theta _2}h)}}{{\left( {n + 1} \right){f^{(n + 1)}}(x + {\theta _3}h)}} = \frac{1}{{n + 1}}\]
