存在正实数$\varepsilon$,$f$是在$(-\varepsilon,\varepsilon)$里的$3$阶可导函数,且3阶导函数是连续的.则
$$f(h)=f(0)+hf'(0)+\frac{h^2}{2!}f''(0)+\frac{h^3}{3!}f'''(0)+o(h^3)$$其中$$\lim_{h\to 0}\frac{o(h^3)}{h^3}=0$$ 在证明这个式子之前，先说明这个式子是怎么发现的.在历史上，泰勒公式是利用多项式插值而发现的.
当$h>0$时，将区间$[0,h]$三等分，每段等分的长度为$\Delta x=\frac{h}{3}$.三个等分点分别为$0,\Delta x,2\Delta x,3\Delta x$.根据,经过$$(0,f(0)),(\Delta x,f(\Delta x)),(2\Delta x,f(2\Delta x)),(3\Delta x,f(3\Delta x))$$的多项式为
\begin{align*}
k(x)=f(0)+\frac{x}{1}\frac{\Delta y_0}{\Delta x}+\frac{x(x-\Delta x)}{1\cdot 2}\frac{\Delta^2y_0}{(\Delta x)^2}+\frac{x(x-\Delta x)(x-2\Delta x)}{1\cdot 2\cdot 3}\frac{\Delta^3y_0}{(\Delta x)^3}\end{align*} \begin{equation} \label{eq:30.15.46} \lim_{\Delta x\to 0;\Delta x\neq 0}\frac{\Delta y_0}{\Delta x}=f'(0)\end{equation} \begin{equation} \label{eq:30.15.48} \lim_{\Delta x\to 0;\Delta x\neq 0} \frac{\Delta^2y_0}{(\Delta x)^2}=\lim_{\Delta x\to 0;\Delta x\neq 0}\frac{\Delta y_1-\Delta y_0}{(\Delta x)^2}\end{equation} \begin{equation} \label{eq:30.17.02} \lim_{\Delta x\to 0;\Delta x\neq 0} \frac{\Delta y_1-\Delta y_0}{(\Delta x)^2}=\lim_{\Delta x\to 0;\Delta x\neq 0}\frac{(y_2-y_1)-(y_1-y_0)}{(\Delta x)^2}\end{equation}根据洛必达法则，
\begin{align*}
\lim_{\Delta x\to 0;\Delta x\neq 0}\frac{f(2\Delta x)-2f(\Delta x)+f(0)}{(\Delta x)^2}=\lim_{\Delta x\to 0;\Delta x\neq 0}\frac{f'(2\Delta x)-f'(\Delta x)}{\Delta x}=\lim_{\Delta x\to 0;\Delta x\neq 0}[2f''(2\Delta x)-f''(\Delta x)]\end{align*}而$$\lim_{\Delta x\to 0;\Delta x\neq 0}[2f''(2\Delta x)-f''(\Delta
x)]=f''(0)$$.
下面看
\begin{equation} \label{eq:31.19.36} \lim_{\Delta x\to 0;\Delta x\neq 0}\frac{\Delta^3y_0}{(\Delta x)^3}\end{equation}.\begin{equation}
( \Delta^3 y_0)=\Delta^2y_1-\Delta^2y_0=(\Delta y_2-\Delta y_1)-(\Delta y_1-\Delta y_0)\end{equation}因此
\begin{align*} \lim_{\Delta x\to 0;\Delta x\neq 0}\frac{\Delta^3y_0}{(\Delta x)^3}=\lim_{\Delta x\to 0;\Delta x\neq 0}\frac{(\Delta y_2-\Delta y_1)-(\Delta y_1-\Delta y_0)}{(\Delta x)^3}=\lim_{\Delta x\to 0;\Delta x\neq 0}\frac{y_3-3y_2+3y_1-y_0}{(\Delta x)^3}\end{align*}根据洛必达法则，
\begin{align*} \lim_{\Delta x\to 0;\Delta x\neq 0}\frac{y_3-3y_2+3y_1-y_0}{(\Delta x)^3}&=\lim_{\Delta x\to 0;\Delta x\neq 0}\frac{3f'(3\Delta x)-6f'(2\Delta x)+3f'(\Delta x)}{3(\Delta x)^2}\\&=\lim_{\Delta x\to 0;\Delta x\neq 0}\frac{9f''(3\Delta x)-12f''(2\Delta x)+3f''(\Delta x)}{6\Delta x}\\&=\lim_{\Delta x\to 0;\Delta x\neq 0}\frac{27f'''(3\Delta x)-24f'''(2\Delta x)+3f'''(\Delta x)}{6}\end{align*} 因为\begin{align*}\begin{cases}\lim_{\Delta x\to 0;\Delta x\neq 0}f'''(3\Delta x)=f'''(0)\\\lim_{\Delta x\to 0;\Delta x\neq 0}f'''(2\Delta x)=f'''(0)\\\lim_{\Delta x\to 0;\Delta x\neq 0}f'''(\Delta x)=f'''(0)\\\end{cases}\end{align*}(根据三阶导函数仍连续)因此
\begin{equation} \label{eq:1.12.31} \lim_{\Delta x\to 0;\Delta x\neq 0}\frac{y_3-3y_2+3y_1-y_0}{(\Delta x)^3}=f'''(0)\end{equation} 综上可见，当$h$足够小时，$\Delta x$也足够小，此时，\begin{equation}
\label{eq:1.12.32} k(x)\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3\end{equation}
上面谈了发现过程,下面严格证明该命题.我们证明
\begin{equation}
\label{eq:1.12.46} \lim_{h\to 0}\frac{f(h)-[f(0)+f'(0)h+\frac{f''(0)}{2!}h^2+\frac{f'''(0)}{3!}h^3]}{h^3}=0\end{equation}由洛必达法则，即证
$$\lim_{h\to 0}\frac{f'(h)-f'(0)-hf''(0)-(h^2/2)f'''(0)}{3h^2}=0$$即证
$$\lim_{h\to 0}\frac{f''(h)-f''(0)-hf'''(0)}{6h}=0$$即证
$$\frac{1}{6}\lim_{h\to 0}(\frac{f''(h)-f''(0)}{h}-f'''(0))=0$$而根据三阶导函数的连续性，这是成立的.