题目：设 $f(x)$ 三阶可导, 且 $f^{\prime \prime \prime}(a) \neq 0$, \(f(x)=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}[a+\theta(x-a)](x-a)^{2}}{2} \quad(0<\theta<1),\) 求证: $\lim _{x \rightarrow a} \theta=\frac{1}{3}$.
证明：
把 $f(x)$ 在 $x=a$ 处展为泰勒公式: \(\begin{aligned} &f(x)=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}+\frac{f^{\prime \prime \prime}(a)}{3 !}(x-a)^{3}+o\left((x-a)^{3}\right), \\ &f^{\prime \prime}(x)=f^{\prime \prime}(a)+f^{\prime \prime \prime}(a)(x-a)+o(x-a) . \end{aligned}\) 把 $x=a+\theta(x-a)$ 代入上式得 : $f^{\prime \prime}[a+\theta(x-a)]=f^{\prime \prime}(a)+f^{\prime \prime \prime}(a) \theta(x-a)+o(x-a) .$
另一方面:
$f^{\prime \prime}[a+\theta(x-a)]=f^{\prime \prime}(a)+\frac{f^{\prime \prime \prime}(a)}{3}(x-a)+o(x-a)$. (联立得:
$\frac{1}{3} f^{\prime \prime \prime}(a)(x-a)+o(x-a)=f^{\prime \prime \prime}(a) \theta(x-a)+o(x-a)$. 所以 $\lim _{x \rightarrow a} \theta=\frac{1}{3}$.