P317练习3

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--- # P317练习3 设 $\displaystyle x\neq 0$, $\displaystyle f(x)=\int\_0^x \sin\frac\{1\}\{t\}\mathrm\{ d\} t$, $\displaystyle f(0)=0$, 求证: $\displaystyle f'(0)=0$. (北京大学习题) [购书](https://mp.weixin.qq.com/s/9RUc7PhEAjT7vlZk4lqZSw) / [答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A) / [pdf1](https://mbd.pub/o/zhangzujin) / [pdf2](https://mbd.pub/o/gjsx) / \begin\{aligned\} \mbox\{原式\}=&\lim\_\{x\to 0\}\frac\{f(x)-f(0)\}\{x-0\} =\lim\_\{x\to 0\} \int\_0^x t^2\cdot \frac\{1\}\{t^2\}\sin\frac\{1\}\{t\}\mathrm\{ d\} t =\lim\_\{x\to 0\}\frac\{1\}\{x\}\int\_0^x t^2\mathrm\{ d\} \cos\frac\{1\}\{t\}\\\\ &=\lim\_\{x\to 0\}\frac\{1\}\{x\} \left\[\left.t^2\cos\frac\{1\}\{t\}\right|\_0^x -\int\_0^x 2t\cos\frac\{1\}\{t\} \mathrm\{ d\} t\right\] =\lim\_\{x\to 0\}\frac\{1\}\{x\}\left\[x^2\cos\frac\{1\}\{x\} -2\int\_0^x t\cos\frac\{1\}\{t\}\mathrm\{ d\} t\right\]\\\\ &\xlongequal[\tiny\mbox\{分中值\}]\{\tiny\mbox\{第一积\}\}\lim\_\{x\to 0\}\left\[x\cos \frac\{1\}\{x\}-\frac\{2\}\{x\}\cos\frac\{1\}\{\xi\_x\} \int\_0^x t\mathrm\{ d\} t\right\] =\lim\_\{x\to 0\}\left\[x\cos\frac\{1\}\{x\}-x\cos\frac\{1\}\{\xi\_x\}\right\] \xlongequal[\tiny\mbox\{无穷小\}]\{\tiny\mbox\{有界 $\displaystyle \cdot$\}\} 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw) / [公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow) / [资料目录](https://mbd.pub/o/bread/mbd-YpiZmJdv) / [视频](https://space.bilibili.com/507709073) / [微信群](https://mbd.pub/o/bread/mbd-YZmTkp1w)