上册 3.4 与导数有关的极限 第11题

\section*{解题过程：} （1）由拉格朗日中值定理，$\displaystyle \sqrt{x+1}-\sqrt{x}=\frac{1}{2 \sqrt{x+\theta(x)}}$ ．即 $$ \theta(x)=\frac{1}{4}+\frac{\sqrt{x^{2}+x}-x}{2}=\frac{1}{4}+\frac{x}{2\left(\sqrt{x^{2}+x}+x\right)} $$ 所以 $\displaystyle \frac{1}{4} \leqslant \theta(x) \leqslant \frac{1}{2}$ ． $$ \lim _{x \rightarrow 0^{+}} \theta(x)=\lim _{x \rightarrow 0^{+}}\left[\frac{1}{4}+\frac{x}{2\left(\sqrt{x^{2}+x}+x\right)}\right]=\frac{1}{4} ; \lim _{x \rightarrow+\infty} \theta(x)=\lim _{x \rightarrow+\infty}\left[\frac{1}{4}+\frac{x}{2\left(\sqrt{x^{2}+x}+x\right)}\right]=\frac{1}{4}+\frac{1}{4}=\frac{1}{2} . $$ （2）由泰勒公式知 $$ f(a+h)=f(a)+f^{\prime}(a) h+\frac{f^{\prime \prime}(a)}{2!} h^{2}+o\left(h^{2}\right) $$ 与已知等式比较整理得 $$ f^{\prime}(a+\theta h)-f^{\prime}(a)=\frac{f^{\prime \prime}(a)}{2} h+o(h) . $$ 由拉格朗日定理得 $$ f^{\prime \prime}\left(a+\theta_{1} \theta h\right) \theta h=\frac{f^{\prime \prime}(a)}{2} h+o(h) \text {, 即 } \theta=\frac{1}{2} \cdot \frac{f^{\prime \prime}(a)}{f^{\prime \prime}\left(a+\theta_{1} \theta h\right)}+\frac{o(h)}{f^{\prime \prime}\left(a+\theta_{1} \theta h\right)} \text {. } $$ 所以 $\displaystyle \lim _{h \rightarrow 0} \theta=\frac{1}{2}$ ． （3）由题设 $$ f(x+h)=f(x)+f^{\prime}(x) h+\cdots+\frac{h^{n}}{n!} f^{(n)}(x+\theta h), 0<\theta<1 $$ 由 Taylor 公式得 $\displaystyle f(x+h)=f(x)+f^{\prime}(x) h+\cdots+\frac{h^{n}}{n!} f^{(n)}(x)+\frac{h^{n+1}}{(n+1)!} f^{(n+1)}(x)+o\left(h^{n+1}\right)$ ． 比较得 $$ f^{(n)}(x+\theta h)-f^{(n)}(x)=\frac{h}{n+1} f^{(n+1)}(x)+o(h), $$ 即 $$ \theta \cdot \frac{f^{(n)}(x+\theta h)-f^{(n)}(x)}{\theta h}=\frac{f^{(n+1)}(x)}{n+1}+\frac{o(h)}{h} . $$ 令 $h \rightarrow 0$ 得 $\displaystyle \theta f^{(n+1)}(x)=\frac{f^{(n+1)}(x)}{n+1}$ ．故 $\displaystyle \lim _{h \rightarrow 0} \theta=\frac{1}{n+1}$ ． （4）由 Taylor 公式 $$ f(a+h)=f(a)+f^{\prime}(a) h+\cdots+\frac{f^{(n)}(a)}{n!} h^{n}+o\left(h^{n}\right)=f(a)+f^{\prime}(a) h+\frac{f^{(n)}(a)}{n!} h^{n}+o\left(h^{n}\right) $$ 与已知等式比较整理得 $$ f^{\prime}(a+\theta h)=f^{\prime}(a)+\frac{f^{(n)}(a)}{n!} h^{n-1}+o\left(h^{n-1}\right) $$ 两边对 $h$ 求 $n-1$ 阶导数得 $$ f^{(n)}(a+\theta h) \theta^{n-1}=\frac{f^{(n)}(a)}{n}+o(h) . $$ 所以 $\displaystyle \theta^{n-1}=\frac{1}{n} \cdot \frac{f^{(n)}(a)}{f^{(n)}(a+\theta h)}+\frac{o(h)}{f^{(n)}(a+\theta h)}$ ．故 $\displaystyle \lim _{h \rightarrow 0} \theta^{n-1}=\frac{1}{n}$ ，即 $\displaystyle \lim _{h \rightarrow 0} \theta=n^{-\frac{1}{n-1}}$ ． （5）由中值定理，存在 $\xi \in(0, x)$ 使得 $\int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d} t=\mathrm{e}^{\xi^{2}} x$ ．对 $\xi^{2} \in\left(0, x^{2}\right)$ ，存在 $\theta \in(0,1)$ 使得 $\xi^{2}=\theta x^{2}$ ． 故 $$ \int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d} t=\mathrm{e}^{\theta x^{2}} x $$ 若存在 $\theta_{1}, \theta_{2} \in(0,1)$ ，使 $\int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d} t=x \mathrm{e}^{\theta_{1} x^{2}}, \int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d} t=x \mathrm{e}^{\theta_{2} x^{2}}$ ，则 $\theta_{1}=\theta_{2}$ 。于是存在唯一 $\theta \in(0,1)$ 使得 $$ \int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d} t=x \mathrm{e}^{\theta x^{2}} \text {, 且 } \theta(x)=\frac{1}{x^{2}} \ln \frac{\int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d} t}{x}=\frac{\ln \int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d} t-\ln x}{x^{2}}=\frac{\ln \int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d} t}{x^{2}}-\frac{\ln x}{x^{2}} \text {. } $$ 因 $$ \lim _{x \rightarrow+\infty} \frac{\ln \int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d} t}{x^{2}}=\lim _{x \rightarrow+\infty} \frac{\mathrm{e}^{x^{2}}}{2 x \int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d} t}=\frac{1}{2} \lim _{x \rightarrow+\infty} \frac{\frac{1}{x} \mathrm{e}^{x^{2}}}{\int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d} t}=\frac{1}{2} \lim _{x \rightarrow+\infty} \frac{-\frac{1}{x^{2}} \mathrm{e}^{x^{2}}+2 \mathrm{e}^{x^{2}}}{\mathrm{e}^{x^{2}}}=1, \lim _{x \rightarrow \infty} \frac{\ln x}{x^{2}}=0, $$ 故 $\lim _{x \rightarrow+\infty} \theta(x)=1$ ． （6）证明与（5）类似． （7）由已知得 $f\left(\theta_{n}\right)=\sqrt[n]{\left(\int_{0}^{1} f^{n}(x) \mathrm{d} x\right)}, \lim _{n \rightarrow \infty} f\left(\theta_{n}\right)=\lim _{n \rightarrow x} \sqrt[n]{\left(\int_{0}^{1} f^{n}(x) \mathrm{d} x\right)}=\max _{[0,1]} f(x)=f(1)$ 。所以 $\lim _{n \rightarrow+\infty} \theta_{n}=1$.