## 张祖锦2023年数学专业真题分类70天之第09天
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185、 6、 (12 分) 设 $\displaystyle f(x)$ 在 $\displaystyle [1,+\infty)$ 上有定义, 且存在正常数 $\displaystyle l,L$, 对任意的 $\displaystyle x\_1,x\_2\in [1,+\infty)$, 都有
\begin\{aligned\} l|x\_2-x\_1|\leq |f(x\_2)-f(x\_1)|\leq L|x\_2-x\_1|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
证明: 存在 $\displaystyle X\in [1,+\infty)$, 使得 $\displaystyle \frac\{x+\mathrm\{e\}^\{-x\}\}\{f(x)\}$ 在 $\displaystyle [X,+\infty)$ 上一致连续. (吉林大学2023年数学分析考研试题) [连续 ]
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https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由题设,
\begin\{aligned\} x > 1\Rightarrow&l(x-1)\leq |f(x)-f(1)|\leq L(x-1)\\\\ \Rightarrow&l(x-1)-|f(1)|\leq |f(x)|\leq L(x-1)+|f(1)|\\\\ \Rightarrow& l\left(1-\frac\{1\}\{x\}\right)-\frac\{|f(1)|\}\{x\} \leq \frac\{|f(x)|\}\{x\}\leq L\left(1-\frac\{1\}\{x\}\right)+\frac\{|f(1)|\}\{x\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
从而存在 $\displaystyle X > 1$, 使得
\begin\{aligned\} \forall\ x\geq X, |f(x)|\geq 1, \frac\{l\}\{2\} < \frac\{|f(x)|\}\{x\} < \frac\{3L\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由
\begin\{aligned\} &\left|\frac\{x\}\{f(x)\}-\frac\{y\}\{f(y)\}\right| \leq \left|\frac\{x-y\}\{f(x)\}\right| +|y|\cdot \left|\frac\{1\}\{f(x)\}-\frac\{1\}\{f(y)\}\right|\\\\ \leq&|x-y|+\frac\{|y|\}\{|f(y)|\}\cdot \frac\{1\}\{|f(x)|\}|f(x)-f(y)|\\\\ \leq&|x-y|+\frac\{2\}\{3L\}\cdot \frac\{1\}\{1\}\cdot L|x-y| < 2|x-y|, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
\begin\{aligned\} &\left|\frac\{\mathrm\{e\}^\{-x\}\}\{f(x)\}-\frac\{\mathrm\{e\}^\{-y\}\}\{f(y)\}\right| \leq \left|\frac\{\mathrm\{e\}^\{-x\}-\mathrm\{e\}^\{-y\}\}\{f(x)\}\right| \mathrm\{e\}^\{-y\}\left|\frac\{1\}\{f(x)\}-\frac\{1\}\{f(y)\}\right|\\\\ \xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\}&\frac\{\mathrm\{e\}^\{-\xi\_\{xy\}\}|x-y|\}\{|f(x)|\}+\mathrm\{e\}^\{-y\}\frac\{|f(x)-f(y)|\}\{|f(x)f(y)|\}\\\\ \leq& |x-y|+L|x-y|=(1+L)|x-y| \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知设 $\displaystyle F(x)=\frac\{x+\mathrm\{e\}^\{-x\}\}\{f(x)\}$ 后,
\begin\{aligned\} \forall\ \varepsilon > 0,\exists\ \delta=\frac\{\varepsilon\}\{L+3\} > 0,\mathrm\{ s.t.\} \forall\ x,y\geq X, |x-y| < \delta, |f(x)-f(y)| < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而 $\displaystyle F(x)=\frac\{x+\mathrm\{e\}^\{-x\}\}\{f(x)\}$ 在 $\displaystyle [X,+\infty)$ 上一致连续.跟锦数学微信公众号. [在线资料](
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186、 2、 对任意的正数 $\displaystyle A$, 证明: $\displaystyle f(x)=\sin x^2$ 在 $\displaystyle [0,A]$ 上一致连续, 但在 $\displaystyle \mathbb\{R\}$ 上非一致连续. (南京航空航天大学2023年数学分析考研试题) [连续 ]
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle f(x)$ 在 $\displaystyle [0,A]$ 上连续, 而一致连续. 又由
\begin\{aligned\} &x\_n=\sqrt\{2n\pi+\frac\{\pi\}\{2\}\}, y\_n=\sqrt\{2n\pi\}\\\\ \Rightarrow& \lim\_\{n\to\infty\}(x\_n-y\_n)=\lim\_\{n\to\infty\}\frac\{\frac\{\pi\}\{2\}\}\{x\_n+y\_n\}=0, |f(x\_n)-f(y\_n)|=1 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle f$ 在 $\displaystyle \mathbb\{R\}$ 上非一致连续.跟锦数学微信公众号. [在线资料](
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187、 4、 解答如下问题. (1)、 用定义证明 $\displaystyle f(x)=\frac\{1\}\{x\}$ 在 $\displaystyle [1,+\infty)$ 上一致连续, 在 $\displaystyle (0,1)$ 上非一致连续. (南京师范大学2023年数学分析考研试题) [连续 ]
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle \forall\ \varepsilon > 0,\exists\ \delta=\varepsilon > 0,\mathrm\{ s.t.\} \forall\ x,y\geq 1$,
\begin\{aligned\} |x-y| < \delta, |f(x)-f(y)|=\frac\{|x-y|\}\{xy\} \leq |x-y| < \varepsilon \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle f(x)=\frac\{1\}\{x\}$ 在 $\displaystyle [1,+\infty)$ 上一致连续. 又由 $\displaystyle \exists\ \varepsilon\_0=1, \forall\ \delta > 0,\exists\ n > \frac\{1\}\{2\delta\}$,
\begin\{aligned\} x=\frac\{1\}\{n\}, y=\frac\{1\}\{2n\}, |x-y|=\frac\{1\}\{2n\} < \delta, |f(x)-f(y)|=n\geq \varepsilon\_0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle f(x)=\frac\{1\}\{x\}$ 在 $\displaystyle (0,1)$ 上非一致连续.跟锦数学微信公众号. [在线资料](
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188、 5、 (20 分) 设 $\displaystyle \alpha$ 为实数,
\begin\{aligned\} f(x)=\left\\{\begin\{array\}\{llllllllllll\}x^\alpha\cos\frac\{1\}\{x\},&x > 0,\\\\ 0,&x=0.\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
已知 $\displaystyle f(x)$ 在 $\displaystyle [0,+\infty)$ 上一致连续, 求 $\displaystyle \alpha$ 的范围. (南开大学2023年数学分析考研试题) [连续 ]
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle \alpha$ 的取值范围为 $\displaystyle (0,1]$. 事实上, (1)、 由题设知 $\displaystyle f$ 在 $\displaystyle x=0$ 处连续,
\begin\{aligned\} 0=f(0)=\lim\_\{x\to 0\}f(x) \xlongequal[\tiny\mbox\{原理\}]\{\tiny\mbox\{归结\}\} \lim\_\{n\to\infty\}f\left(\frac\{1\}\{2n\pi\}\right)=\lim\_\{n\to\infty\}\frac\{1\}\{(2n\pi)^\alpha\}\Rightarrow \alpha > 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 我们先给出一个结论. 设函数 $\displaystyle f(x)$ 在区间 $\displaystyle [1,+\infty)$ 一致连续. 则函数 $\displaystyle g(x)=\frac\{f(x)\}\{x\}$ 在区间 $\displaystyle [1,+\infty)$ 上有界. 事实上, 由函数 $\displaystyle f(x)$ 在 $\displaystyle [1,+\infty)$ 上一致连续知
\begin\{aligned\} \exists\ \delta > 0,\mathrm\{ s.t.\} \forall\ x,y\in [1,+\infty), |x-y|\leq\delta, |f(x)-f(y)| < 1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则对 $\displaystyle \forall\ x\geq 1$,
\begin\{aligned\} \exists\ |\ n,\mathrm\{ s.t.\} n\delta\leq x-1 < (n+1)\delta \Rightarrow 1+n\delta\leq x < 1+(n+1)\delta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是
\begin\{aligned\} |f(x)|&\leq |f(x)-f(1+n\delta)| +\sum\_\{k=1\}^n \left|f(1+k\delta)-f\left(1+(k-1)\delta\right)\right|\\\\ &\leq 1+\sum\_\{k=1\}^n 1=n+1 < \frac\{x-1\}\{\delta\}+1 < \frac\{x\}\{\delta\}+1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故
\begin\{aligned\} \left|\frac\{f(x)\}\{x\}\right|\leq \frac\{1\}\{\delta\}+\frac\{1\}\{x\}\leq \frac\{1\}\{\delta\}+1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(3)、 回到题目. 若 $\displaystyle \alpha > 1$, 则由
\begin\{aligned\} \lim\_\{x\to+\infty\}\frac\{f(x)\}\{x\}=\lim\_\{x\to+\infty\}x^\{\alpha-1\}\cos\frac\{1\}\{x\}=\lim\_\{x\to+\infty\}x^\{\alpha-1\}=+\infty \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及第 2 步的逆否命题知 $\displaystyle f$ 不一致连续. (4)、 当 $\displaystyle 0 < \alpha\leq 1$ 时, 由
\begin\{aligned\} x > 0\Rightarrow& f'(x)=\alpha x^\{\alpha-1\}\cos\frac\{1\}\{x\}s+x^\{\alpha-2\}\sin\frac\{1\}\{x\}\\\\ \Rightarrow&\lim\_\{x\to+\infty\}f'(x)\xlongequal[\tiny\mbox\{无穷小\}]\{\tiny\mbox\{有界 $\displaystyle \cdot$\}\} \left\\{\begin\{array\}\{llllllllllll\}0,&0 < \alpha < 1,\\\\ 1,&\alpha=1\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} \exists\ X > 0,\mathrm\{ s.t.\} \forall\ x\geq X, |f'(x)|\leq 2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
从而
\begin\{aligned\} &\forall\ \varepsilon > 0,\exists\ \delta=\frac\{\varepsilon\}\{2\}, \mathrm\{ s.t.\}\forall\ X\leq x < y < x+\delta,\\\\ &|f(x)-f(y)|\xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\} |f'(\xi)|\cdot |y-x|\leq 2|y-x| < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle f$ 在 $\displaystyle [X,+\infty)$ 上一致连续. 又由 $\displaystyle f$ 在 $\displaystyle [0,X]$ 上连续知 $\displaystyle f$ 在 $\displaystyle [0,+\infty)$ 上一致连续.跟锦数学微信公众号. [在线资料](
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189、 2、 $\displaystyle f(x),g(x)$ 在区间 $\displaystyle I$ 上有界且一致连续. 证明: $\displaystyle f(x)g(x)$ 在 $\displaystyle I$ 上一致连续. (山西大学2023年数学分析考研试题) [连续 ]
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https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle f,g$ 在 $\displaystyle I$ 上有界知
\begin\{aligned\} \exists\ M > 0,\mathrm\{ s.t.\} \forall\ x\in I, |f(x)|\leq M, |g(x)|\leq M. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
又由 $\displaystyle f,g$ 在区间 $\displaystyle I$ 上一致连续知对 $\displaystyle \forall\ \varepsilon > 0$,
\begin\{aligned\} \exists\ \delta\_1 > 0,\mathrm\{ s.t.\} x,y\in I,\ |x-y| < \delta\_1&\Rightarrow |f(x)-f(y)| < \frac\{\varepsilon\}\{2M\},\\\\ \exists\ \delta\_2 > 0,\mathrm\{ s.t.\} x,y\in I,\ |x-y| < \delta\_2&\Rightarrow |g(x)-g(y)| < \frac\{\varepsilon\}\{2M\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
取 $\displaystyle \delta=\min\left\\{\delta\_1,\delta\_2\right\\} > 0$, 则当 $\displaystyle x,y\in I,\ |x-y| < \delta$ 时,
\begin\{aligned\} \left|f(x)g(x)-f(y)g(y)\right|&=\left|f(x)[g(x)-g(y)]+[f(x)-f(y)]g(y)\right|\\\\ & < M\cdot \frac\{\varepsilon\}\{2M\}+\frac\{\varepsilon\}\{2M\}\cdot M=\varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这表明 $\displaystyle f\cdot g$ 在 $\displaystyle I$ 上一致连续.跟锦数学微信公众号. [在线资料](
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190、 7、 (20 分) 已知 $\displaystyle f(x)$ 在 $\displaystyle (-\infty,+\infty)$ 上连续, 且 $\displaystyle \lim\_\{x\to-\infty\}f(x)$ 和 $\displaystyle \lim\_\{x\to+\infty\}f(x)$ 都存在. 证明: $\displaystyle f(x)$ 在 $\displaystyle (-\infty,+\infty)$ 上一致连续. (陕西师范大学2023年数学分析考研试题) [连续 ]
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https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle \lim\_\{x\to-\infty\}f(x)=A, \lim\_\{x\to+\infty\}f(x)=B$, 则由 Cauchy 收敛准则知
\begin\{aligned\} \forall\ \varepsilon > 0,\exists\ X > 0,\mathrm\{ s.t.\} \left\\{\begin\{array\}\{llllllllllll\}x,x'\geq X\Rightarrow |f(x)-f(x')| < \varepsilon,\\\\ x,x'\leq -X\Rightarrow |f(x)-f(x')| < \varepsilon.\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
又由 $\displaystyle f$ 在 $\displaystyle [-X-1,X+1]$ 上连续知其一致连续, 而
\begin\{aligned\} \exists\ \delta\in(0,1),\mathrm\{ s.t.\}\forall\ x,x'\in [-X-1,X+1], |x-x'| < \delta, |f(x)-f(x')| < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是对 $\displaystyle \forall\ x,x'\in\mathbb\{R\}, x < x' < x+\delta$, (1)、 若 $\displaystyle x < -X-1$, 则 $\displaystyle x' < -X$; (2)、 若 $\displaystyle -X-1\leq x < X$, 则 $\displaystyle -X-1\leq x < x' < X+1$; (3)、 若 $\displaystyle x\geq X$, 则 $\displaystyle x'\geq X$. 不论何种情形, 都有 $\displaystyle |f(x)-f(x')| < \varepsilon$. 故 $\displaystyle f$ 在 $\displaystyle \mathbb\{R\}$ 上一致连续.跟锦数学微信公众号. [在线资料](
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191、 4、 解答如下问题: (1)、 叙述 $\displaystyle y=f(x)$ 在区间 $\displaystyle I$ 上一致连续的定义; (2)、 讨论 $\displaystyle y=\tan x$ 在区间 $\displaystyle \left(0,\frac\{\pi\}\{4\}\right)$ 与 $\displaystyle \left(0,\frac\{\pi\}\{2\}\right)$ 的一致连续性, 并证明之. (上海大学2023年数学分析考研试题) [连续 ]
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https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 若
\begin\{aligned\} \forall\varepsilon > 0,\exists\ \delta > 0,\mathrm\{ s.t.\} \forall\ x,x'\in I, |x-x'| < \delta, |f(x)-f(x')| < \varepsilon, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则称 $\displaystyle f$ 在 $\displaystyle I$ 上一致连续. 对 $\displaystyle y=f(x)=\tan x$, 由
\begin\{aligned\} &\forall\ \varepsilon > 0,\exists\ \delta=\frac\{\varepsilon\}\{2\},\mathrm\{ s.t.\} \forall\ x', x''\in \left(0,\frac\{\pi\}\{4\}\right),\\\\ &|x'-x''| < \delta, |f(x')-f(x'')|\xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\} |\sec^2\xi| |x'-x''| < 2\delta =\varepsilon \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle f$ 在 $\displaystyle \left(0,\frac\{\pi\}\{4\}\right)$ 上一致连续. 又由
\begin\{aligned\} &\left|f\left(\frac\{\pi\}\{-\delta\}\right)-f\left(\frac\{\pi\}\{2\}-2\delta\right)\right| =|\cot \delta-\cot 2\delta|\\\\ =&\left|\frac\{\sin \delta\}\{\cos \delta\}-\frac\{\sin 2\delta\}\{\cos 2\delta\}\right| =\frac\{\sin \delta\}\{\sin \delta\sin 2\delta\}=\frac\{1\}\{\sin 2\delta\}\xrightarrow\{\delta\to 0^+\}+\infty \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle f$ 在 $\displaystyle \left(0,\frac\{\pi\}\{2\}\right)$ 上不一致连续.跟锦数学微信公众号. [在线资料](
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192、 1、 是非判断题. 判断下列说法是否正确, 正确的打’$\surd$‘, 错误的打’$\times$‘. 每小题 5 分, 共 25 分. (1)、 存在处处不连续但有界的函数. (上海交通大学2023年数学分析考研试题) [连续 ]
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle \surd$. 比如 Dirichlet 函数.跟锦数学微信公众号. [在线资料](
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193、 9、 (本题 15 分) 设 $\displaystyle f(x)$ 在 $\displaystyle [2,+\infty)$ 上满足 Lipschitz 条件, 即存在 $\displaystyle k > 0$, 使得
\begin\{aligned\} |f(x)-f(y)|\leq k|x-y|, \forall\ x,y\in [2,+\infty). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
证明: $\displaystyle \frac\{f(x)\}\{x\}$ 在 $\displaystyle [2,+\infty)$ 上一致连续. (四川大学2023年数学分析考研试题) [连续 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由
\begin\{aligned\} \left|\frac\{f(x)\}\{x\}\right|&\leq \frac\{|f(x)-f(2)|+|f(2)|\}\{x\} \leq\frac\{k(x-2)+|f(2)|\}\{x\} \leq k+|f(2)| \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} &\quad \left|\frac\{f(x)\}\{x\}-\frac\{f(y)\}\{y\}\right| =\left|\frac\{yf(x)-xf(y)\}\{xy\}\right| =\frac\{|y[f(x)-f(y)]+(y-x)f(y)|\}\{xy\}\\\\ &\leq \frac\{1\}\{x\} k|x-y| +\frac\{1\}\{x\}\frac\{|f(y)|\}\{y\} |y-x| \leq k|x-y|+(k+|f(2)|)|y-x|\\\\ &\leq (2k+|f(2)|)|x-y|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故
\begin\{aligned\} &\forall\ \varepsilon > 0,\exists\ \delta=\frac\{\varepsilon\}\{2k+|f(2)|\} > 0,\mathrm\{ s.t.\}\forall\ x,y\geq 2, |x-y| < \delta,\\\\ &\left|\frac\{f(x)\}\{x\}-\frac\{f(y)\}\{y\}\right|\leq (2k+|f(2)|)|x-y| < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这就证明了 $\displaystyle \frac\{f(x)\}\{x\}$ 在 $\displaystyle [2,+\infty)$ 上一致连续.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](
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---
194、 6、 (15 分) 设 $\displaystyle \alpha$ 为实数, 讨论 $\displaystyle f(x)=x^\alpha$ 在 $\displaystyle (1,+\infty)$ 上的一致连续性. (苏州大学2023年数学分析考研试题) [连续 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 当 $\displaystyle \alpha\leq 0$ 时, $\displaystyle \lim\_\{x\to+\infty\}f(x)=\left\\{\begin\{array\}\{llllllllllll\}0,&\alpha < 0,\\\\ 1,&\alpha=0.\end\{array\}\right.$ 由 Cauchy 收敛准则知
\begin\{aligned\} \{\color\{red\}\forall\ \varepsilon > 0\}, \exists\ R > 1,\mathrm\{ s.t.\} \forall\ x\geq R, x'\geq R, |f(x')-f(x)| < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
又 $\displaystyle f$ 在 $\displaystyle 1\leq x\leq R+1$ 上连续, 而一致连续,
\begin\{aligned\} &\{\color\{red\}\exists\ \delta\in (0,1)\},\mathrm\{ s.t.\} \forall\ 1\leq x\leq R+1, 1\leq x'\leq R+1,\\\\ &|x'-x| < \delta, |f(x')-f(x)| < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故当 $\displaystyle \{\color\{red\}x,x'\in [1,+\infty), |x'-x| < \delta\}$ 时,
\begin\{aligned\} &\left\\{\begin\{array\}\{llllllllllll\} x\leq R\Rightarrow x'\leq x+|x'-x|\leq R+1\\\\ \mbox\{或\}x'\leq R\Rightarrow x\leq R+1\\\\ \mbox\{或\}x > R, x' > R \end\{array\}\right.\\\\ \Rightarrow&\{\color\{red\}|f(x')-f(x)| < \varepsilon\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle f$ 在 $\displaystyle [1,+\infty)$ 上一致连续. (2)、 当 $\displaystyle 0 < \alpha\leq 1$ 时, (2-1)、 先证
\begin\{aligned\} 0 < \alpha\leq 1, x\geq 0, y\geq 0\Rightarrow (x+y)^\alpha \leq x^\alpha+y^\alpha. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
事实上, 若 $\displaystyle x=0$ 或 $\displaystyle y=0$, 则结论自明. 若 $\displaystyle x > 0$ 且 $\displaystyle y > 0$, 则由 $\displaystyle a^t\ (0 < a < 1)$ 关于 $\displaystyle t$ 递减知
\begin\{aligned\} \frac\{x^\alpha+y^\alpha\}\{(x+y)^\alpha\}=\left(\frac\{x\}\{x+y\}\right)^\alpha+\left(\frac\{y\}\{x+y\}\right)^\alpha \geq\frac\{x\}\{x+y\}+\frac\{y\}\{x+y\}=1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2-2)、 由第 i 步知当 $\displaystyle 0 < \alpha\leq 1, 1\leq x < y$ 时,
\begin\{aligned\} 0 < y^\alpha-x^\alpha=\left\[x+(y-x)\right\]^\alpha-x^\alpha \leq (y-x)^\alpha. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
从而
\begin\{aligned\} \forall\ \varepsilon > 0,\exists\ \delta=\varepsilon^\frac\{1\}\{\alpha\} > 0,\mathrm\{ s.t.\}\forall\ 1 < x < y < x+\delta, |f(x)-f(y)| < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle f$ 在 $\displaystyle [1,+\infty)$ 上一致连续. (3)、 当 $\displaystyle \alpha > 1$ 时, 设 $\displaystyle x\_n=(n+1)^\frac\{1\}\{\alpha\}, y\_n=n^\frac\{1\}\{\alpha\}$, 则由
\begin\{aligned\} 0 < (n+1)^\frac\{1\}\{\alpha\}-n^\frac\{1\}\{\alpha\} =&n^\alpha \left\[\left(1+\frac\{1\}\{n\}\right)^\{\frac\{1\}\{\alpha\}\}-1\right\] < n^\frac\{1\}\{\alpha\} \cdot \left\[\left(1+\frac\{1\}\{n\}\right)-1\right\]\\\\ =&\frac\{1\}\{n^\{1-\frac\{1\}\{\alpha\}\}\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及迫敛性知 $\displaystyle \lim\_\{n\to\infty\}(x\_n-y\_n)=0$, 但 $\displaystyle |f(x\_n)-f(y\_n)|=1$. 故 $\displaystyle f$ 在 $\displaystyle [1,+\infty)$ 上不一致连续.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](
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---
195、 10、 设 $\displaystyle f(x)$ 在有限开区间 $\displaystyle (a,b)$ 上可导, 且 $\displaystyle \lim\_\{x\to a^+\}f'(x)$ 和 $\displaystyle \lim\_\{x\to b^-\}f'(x)$ 存在. 证明: (1)、 $\displaystyle \lim\_\{x\to a^+\}f(x)$ 和 $\displaystyle \lim\_\{x\to b^-\}f(x)$ 都存在; (2)、 $\displaystyle f(x)$ 在 $\displaystyle (a,b)$ 上一致连续且有界. (太原理工大学2023年数学分析考研试题) [连续 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 由 $\displaystyle f'(a+0)=A$ 存在知 $\displaystyle \exists\ \delta\_1\in \left(0,b-a\right),\mathrm\{ s.t.\}$
\begin\{aligned\} \forall\ a < x < a+\delta\_1, |f'(x)-A| < 1\Rightarrow |f'(x)| < |A|+1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是
\begin\{aligned\} &\forall\ \varepsilon > 0,\exists\ \delta\in\left\\{0,\min\left\\{\delta\_1,\frac\{\varepsilon\}\{|A|+1\}\right\\}\right\\},\mathrm\{ s.t.\} \forall\ x,x'\in (a,a+\delta),\\\\ &|f(x)-f(x')|\xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\} |f'(\xi)|\cdot |x-x'| < (|A|+1)\delta < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由函数极限的 Cauchy 收敛准则即知 $\displaystyle f(a+0)$ 存在. 同理, $\displaystyle f(b-0)$ 存在. (2)、 令 $\displaystyle F(x)=\left\\{\begin\{array\}\{llllllllllll\}f(a+0),&x=a,\\\\ f(x),&a < x < b,\\\\ f(b-0),&x=b,\end\{array\}\right.$ 则 $\displaystyle F$ 在 $\displaystyle [a,b]$ 上连续, 而一致连续且有界. 这表明 $\displaystyle f(x)=F(x)|\_\{(a,b)\}$ 在 $\displaystyle (a,b)$ 上一致连续且有界.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](
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---
196、 5、 (15 分) 设 $\displaystyle f(x)$ 在 $\displaystyle (0,1]$ 上可导, 且 $\displaystyle \lim\_\{x\to 0^+\}\sqrt\{x\}f'(x)=A$. 求证: $\displaystyle f(x)$ 在 $\displaystyle (0,1]$ 上一致连续. (西南大学2023年数学分析考研试题) [连续 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle \lim\_\{x\to0^+\}\sqrt\{x\}f'(x)=A$ 存在知
\begin\{aligned\} \exists\ M > 0,\ \exists\ \delta\_1\in (0,a),\mathrm\{ s.t.\} \forall\ 0 < x < \delta\_1, \left|\sqrt\{x\}f'(x)\right|\leq M. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是
\begin\{aligned\} &\forall\ \varepsilon > 0,\exists\ \delta=\left(\frac\{\varepsilon\}\{2M\}\right)^2 > 0,\mathrm\{ s.t.\} \forall\ 0 < x,y\leq \delta\_1, |x-y| < \delta,\\\\ &\left|\frac\{f(x)-f(y)\}\{\sqrt\{x\}-\sqrt\{y\}\}\right| \xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Cauchy\}\} \left|\frac\{f'(\xi)\}\{\frac\{1\}\{2\sqrt\{\xi\}\}\}\right|\leq 2M\\\\ &\Rightarrow |f(x)-f(y)|\leq 2M\left|\sqrt\{x\}-\sqrt\{y\}\right|\leq 2M \sqrt\{|x-y|\} < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这就证明了 $\displaystyle f$ 在 $\displaystyle (0,\delta\_1]$ 上一致连续. 又由 $\displaystyle f$ 在 $\displaystyle [\delta\_1,1]$ 上连续知其一致连续. 从而 $\displaystyle f$ 在 $\displaystyle (0,1]$ 上一致连续.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](
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---
197、 1、 计算题 (每题 5 分, 共 30 分). (1)、 设 $\displaystyle f(x)=\left\\{\begin\{array\}\{llllllllllll\}1,&x > 0,\\\\ 0,&x\leq 0,\end\{array\}\right.$ $\displaystyle g(x)=\left\\{\begin\{array\}\{llllllllllll\}x-1,&x\geq 1,\\\\ 1-x,&x < 1.\end\{array\}\right.$ 求 $\displaystyle f\left(g(x)\right)$ 的间断点. (云南大学2023年数学分析考研试题) [连续 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由
\begin\{aligned\} x > 1\Rightarrow&g(x)=x-1 > 0\Rightarrow f\left(g(x)\right)=1,\\\\ x=1\Rightarrow&g(x)=x-1=0\Rightarrow f\left(g(x)\right)=0,\\\\ x < 1\Rightarrow&g(x)=1-x > 0\Rightarrow f\left(g(x)\right)=1 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle x=0$ 是 $\displaystyle f\circ g$ 的可去间断点.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](
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---
198、 6、 (10 分) 讨论函数 $\displaystyle f(x)=x^2+\sin x$ 的一致连续性. (郑州大学2023年数学分析考研试题) [连续 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle g(x)=x^2$, 则
\begin\{aligned\} x\_n=n+\frac\{1\}\{n\}, y\_n=n\Rightarrow \lim\_\{n\to\infty\}(x\_n-y\_n)=0, g(x\_n)-g(y\_n)\geq 2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle g$ 在 $\displaystyle \mathbb\{R\}$ 上不一致连续. 又由 $\displaystyle \forall\ \varepsilon > 0, \exists\ \delta=\varepsilon > 0,\mathrm\{ s.t.\}$
\begin\{aligned\} \forall\ x,y\in\mathbb\{R\}, |x-y| < \delta, |\sin x-\sin y|\xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\} |\cos \xi|\cdot |x-y| < \varepsilon \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \sin x$ 在 $\displaystyle \mathbb\{R\}$ 上一致连续. 从而 $\displaystyle f(x)=g(x)+\sin x$ 在 $\displaystyle \mathbb\{R\}$ 上不一致连续 (因为两个一致连续函数的和还是一致连续).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](
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---
199、 2、 设 $\displaystyle f(x)$ 在 $\displaystyle [a,b]$ 上连续, 且对任何 $\displaystyle x\in [a,b]$, 存在 $\displaystyle y\in [a,b]$, 使得
\begin\{aligned\} |f(y)|\leq \frac\{1\}\{2\}|f(x)|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
证明: 存在 $\displaystyle \xi\in [a,b]$, 使得 $\displaystyle f(\xi)=0$. (中国科学院大学2023年数学分析考研试题) [连续 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle m=\min\_\{[a,b]\}f=f(\xi)$, 则由题设,
\begin\{aligned\} \exists\ \eta\in [a,b],\mathrm\{ s.t.\} |f(\eta)|\leq \frac\{1\}\{2\}|f(\xi)|=\frac\{m\}\{2\} \Rightarrow m\leq |f(\eta)|\leq\frac\{m\}\{2\}\Rightarrow 0=m=f(\xi). \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](
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---
200、 1、 求常数 $\displaystyle a$ 和 $\displaystyle b$, 使得函数 $\displaystyle f(x)=\left\\{\begin\{array\}\{llllllllllll\}\frac\{\sqrt\{1-ax\}-1\}\{x\},&x < 0,\\\\ ax+b,&0\leq x\leq 1,\\\\ \arctan \frac\{1\}\{x-1\},&x > 1\end\{array\}\right.$ 在 $\displaystyle (-\infty,+\infty)$ 上连续. (中国矿业大学(徐州)2023年数学分析考研试题) [连续 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由题设,
\begin\{aligned\} b&=f(0+0)=f(0-0)=\lim\_\{x\to 0\}\frac\{-a\}\{\sqrt\{1-ax\}+1\}=-\frac\{a\}\{2\},\\\\ a+b&=f(1-0)=f(1+0)=\lim\_\{x\to 1^+\}\arctan\frac\{1\}\{x-1\}=\frac\{\pi\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
解得 $\displaystyle a=\pi, b=-\frac\{\pi\}\{2\}$.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](
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---
201、 5、 设单调有界函数 $\displaystyle f(x)$ 在 $\displaystyle (a,b)$ 上连续, 证明: $\displaystyle f(x)$ 在 $\displaystyle (a,b)$ 上一致连续. (中国矿业大学(徐州)2023年数学分析考研试题) [连续 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle f$ 单调有界知 $\displaystyle f(a+0), f(b-0)$ 都存在. 令
\begin\{aligned\} F(x)=\left\\{\begin\{array\}\{llllllllllll\}f(a+0),&x=a,\\\\ f(x),&a < x < b,\\\\ f(b-0),&x=b.\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle F$ 在 $\displaystyle [a,b]$ 上连续, 而一致连续. 故 $\displaystyle f=F|\_\{(a,b)\}$ 一致连续.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](
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---
202、 2、 (15 分) 设 $\displaystyle f(x)$ 在区间 $\displaystyle [0,1]$ 上连续, 且 $\displaystyle f(0)=0, f(1)=1$. 证明: $\displaystyle \forall\ n\in\mathbb\{N\}$, 存在 $\displaystyle \xi\_n\in [0,1]$, 使得
\begin\{aligned\} f\left(\xi\_n+\frac\{1\}\{n\}\right)=f(\xi\_n)+\frac\{1\}\{n\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(中国人民大学2023年数学分析考研试题) [连续 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 $\displaystyle n=1$ 时, 取 $\displaystyle \xi\_1=0$ 即可. (2)、 $\displaystyle n\geq 2$ 时, 令 $\displaystyle F(x)=f\left(x+\frac\{1\}\{n\}\right)-f(x)$, 则
\begin\{aligned\} \sum\_\{i=0\}^\{n-1\}F\left(\frac\{i\}\{n\}\right) &=\sum\_\{i=0\}^\{n-1\} \left\[f\left(\frac\{i+1\}\{n\}\right)-f\left(\frac\{i\}\{n\}\right)\right\]\\\\ &=f(1)-f(0) =0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设
\begin\{aligned\} F\left(\frac\{k\}\{n\}\right)=\max\_\{0\leq i\leq n-1\} F\left(\frac\{i\}\{n\}\right),\quad F\left(\frac\{l\}\{n\}\right)=\min\_\{0\leq i\leq n-1\}F\left(\frac\{i\}\{n\}\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则
\begin\{aligned\} \begin\{array\}\{rcl\} &0&\\\\ &||&\\\\ F\left(\frac\{l\}\{n\}\right)\leq&\frac\{1\}\{n\}\sum\_\{i=0\}^\{n-1\}F\left(\frac\{i\}\{n\}\right) &\leq F\left(\frac\{k\}\{n\}\right). \end\{array\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
据连续函数介值定理即知
\begin\{aligned\} \mbox\{ $\displaystyle \exists\ x\_n$ 介于 $\displaystyle \frac\{k\}\{n\},\frac\{l\}\{n\}$ 之间, 使得 $\displaystyle F(x\_n)=0$.\} \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](
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---
203、 3、 (15 分) 设 $\displaystyle f(x)$ 为 $\displaystyle (0,1)$ 上的函数, 若 $\displaystyle \mathrm\{e\}^x f(x)$ 与 $\displaystyle \mathrm\{e\}^\{-f(x)\}$ 在 $\displaystyle (0,1)$ 上单调不减. 求证: (1)、 $\displaystyle f(x)$ 在 $\displaystyle (0,1)$ 上单调递减; (2)、 $\displaystyle f(x)$ 在 $\displaystyle (0,1)$ 上连续. (中南大学2023年数学分析考研试题) [连续 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 对 $\displaystyle \forall\ 0 < x < y < 1$,
\begin\{aligned\} \mathrm\{e\}^\{-f(x)\}\leq \mathrm\{e\}^\{-f(y)\}\stackrel\{\ln\}\{\Rightarrow\}-f(x)\leq -f(y)\Rightarrow f(x)\geq f(y). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle f\searrow, f(x-0), f(x+0)$ 都存在. 再者,
\begin\{aligned\} \mathrm\{e\}^xf(x)\leq \mathrm\{e\}^yf(y), \forall\ 0 < x < y < 1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle y\to x^+$ 得
\begin\{aligned\} \mathrm\{e\}^x f(x)\leq \mathrm\{e\}^x f(x+0),\quad \forall\ 0 < x < 1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle x\to y^-$ 得
\begin\{aligned\} \mathrm\{e\}^y f(y-0)\leq \mathrm\{e\}^y f(y) \Rightarrow f(y-0)\leq f(y),\quad \forall\ 0 < y < 1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故
\begin\{aligned\} f(x)\leq f(x+0)\stackrel\{f\searrow\}\{\leq\}f(x-0)\leq f(x) \Rightarrow f(x)=f(x+0)=f(x-0). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这就证明了 $\displaystyle f$ 在 $\displaystyle x$ 处连续. 由 $\displaystyle x$ 的任意性知 $\displaystyle f$ 在 $\displaystyle (0,1)$ 上连续.跟锦数学微信公众号. [在线资料](
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---
204、 3、 (15 分) 证明: 函数 $\displaystyle f(x)$ 在有界区间 $\displaystyle I$ 上一致连续的充分必要条件是当 $\displaystyle \left\\{a\_n\right\\}$ 是 $\displaystyle I$ 上的任意柯西数列时, $\displaystyle \left\\{f(a\_n)\right\\}$ 也是柯西数列. (重庆大学2023年数学分析考研试题) [连续 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 $\displaystyle \Rightarrow$: 由 $\displaystyle f$ 一致连续知
\begin\{aligned\} &\forall\ \varepsilon > 0,\ \exists\ \delta > 0,\mathrm\{ s.t.\} \forall\ x',x''\in A: \ |x'-x''| < \delta, |f(x')-f(x'')| < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设 $\displaystyle \left\\{a\_n\right\\}$ 为 $\displaystyle A$ 中 Cauchy 列, 则对上述 $\displaystyle \delta > 0$,
\begin\{aligned\} \exists\ N,\mathrm\{ s.t.\} \forall\ n,m\geq N, |a\_n-a\_m| < \delta\Rightarrow |f(a\_n)-f(a\_m)| < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 $\displaystyle \Leftarrow$: 用反证法. 若 $\displaystyle f$ 在 $\displaystyle I$ 上不一致连续, 则
\begin\{aligned\} \exists\ \varepsilon\_0 > 0,\ \exists\ a\_n,b\_n\in I: \ |a\_n-b\_n|\to 0,\mbox\{ 但 \}|f(a\_n)-f(b\_n)|\geq \varepsilon\_0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle I$ 有界知 $\displaystyle \left\\{a\_n\right\\}$ 有子列 $\displaystyle \left\\{a\_\{n\_k\}\right\\}$ 收敛于 $\displaystyle \overline\{I\}$ 中某点 $\displaystyle a\_0$. 因 $\displaystyle |a\_n-b\_n|\to 0$, 而 $\displaystyle b\_\{n\_k\}\to a\_0$. 如此,
\begin\{aligned\} a\_\{n\_1\},b\_\{n\_1\},a\_\{n\_2\},b\_\{n\_2\},\cdots \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
为 $\displaystyle I$ 中 Cauchy 列, 但
\begin\{aligned\} f(a\_\{n\_1\}),f(b\_\{n\_1\}),f(a\_\{n\_2\}),f(b\_\{n\_2\}),\cdots \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
不是 Cauchy 列. 这与充分性假设矛盾. 故有结论.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](
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---
205、 (2)、 证明: $\displaystyle \sin^2\sqrt\{x\}$ 在 $\displaystyle [0,+\infty)$ 上一致连续. (重庆师范大学2023年数学分析考研试题) [连续 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle f(x)=\sin^2\sqrt\{x\}=\frac\{1-\cos 2\sqrt\{x\}\}\{2\}$, 则由
\begin\{aligned\} &\forall\ \varepsilon > 0, \exists\ \delta=\varepsilon^2 > 0,\mathrm\{ s.t.\}\forall\ 0\leq x < x' < x+\delta,\\\\ &|f(x)-f(x')|=\frac\{1\}\{2\}|\cos 2\sqrt\{x\}-\cos 2\sqrt\{x'\}|\\\\ &\xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\} \frac\{1\}\{2\}\left|-\cos \xi\_x(2\sqrt\{x\}-2\sqrt\{x'\})\right| \leq |\sqrt\{x\}-\sqrt\{x'\}|\leq \sqrt\{x'-x\} < \sqrt\{\delta\}=\varepsilon \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle f$ 在 $\displaystyle [0,+\infty)$ 上一致连续.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](
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---
206、 4、 (20 分) 设 $\displaystyle f$ 在 $\displaystyle (a,b)$ 上可导, $\displaystyle x\_1,x\_2\in (a,b), x\_1 < x\_2$. 若 $\displaystyle f'(x\_1)f'(x\_2) < 0$, 证明: 存在 $\displaystyle \xi\in (x\_1,x\_2)$, 使得 $\displaystyle f'(\xi)=0$. (安徽大学2023年高等代数考研试题) [微分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 不妨设 $\displaystyle f'(x\_1) < 0, f'(x\_2) > 0$, 则由函数极限的局部保号性知 $\displaystyle \exists\ \delta\in\left(0,\frac\{x\_2-x\_1\}\{2\}\right),\mathrm\{ s.t.\}$
\begin\{aligned\} \forall\ x\in (x\_1,x\_1+\delta), \frac\{f(x)-f(x\_1)\}\{x-x\_1\} < 0\Rightarrow f(x) < f(x\_1),\\\\ \forall\ x\in (x\_2-\delta,x\_2), \frac\{f(x)-f(x\_2)\}\{x-x\_2\} > 0\Rightarrow f(x) < f(x\_2). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle f$ 在 $\displaystyle [x\_1,x\_2]$ 上的最小值只能在 $\displaystyle (x\_1,x\_2)$ 内某点 $\displaystyle \xi$ 处取得. 由极值条件知 $\displaystyle f'(\xi)=0$.跟锦数学微信公众号. [在线资料](
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---
207、 4、 设
\begin\{aligned\} \sum\_\{n=0\}^\infty a\_n(x-x\_0)^n=f(x), x\in (x\_0-r,x\_0+r), r > 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
试证明: $\displaystyle a\_n=\frac\{f^\{(n)\}(x\_0)\}\{n!\}$. 若 $\displaystyle f(x)$ 在 $\displaystyle (x\_0-r,x\_0+r)$ 内有任意阶导数, 且存在 $\displaystyle M > 0$, 使得对任意 $\displaystyle x\in (x\_0-r,x\_0+r)$ 及正整数 $\displaystyle n$ 均有 $\displaystyle |f^\{(n)\}(x)| < M$. 证明:
\begin\{aligned\} f(x)=\sum\_\{n=0\}^\infty \frac\{f^\{(n)\}(x\_0)\}\{n!\}(x-x\_0)^n, x\in (x\_0-r,x\_0+r). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(北京工业大学2023年数学分析考研试题) [微分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 题中幂级数的收敛半径 $\displaystyle R\geq r$, 从而在 $\displaystyle (x\_0-r,x\_0+r)$ 内可任意阶求导. 两边求 $\displaystyle n$ 次导后令 $\displaystyle x=x\_0$ 即得 $\displaystyle a\_nn!=f^\{(n)\}(x\_0)$. (2)、 由
\begin\{aligned\} \sup\_\{x\in (x\_0-r,x\_0+r)\}\left|\frac\{f^\{(n)\}(x\_0)\}\{n!\}(x-x\_0)^n\right| \leq \sup\_\{x\in (x\_0-r,x\_0+r)\}\frac\{M\}\{n!\}|x-x\_0|^n \leq \frac\{M\delta^n\}\{n!\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
$\displaystyle \sum\_\{n=1\}^\infty \frac\{M\delta^n\}\{n!\} < \infty$ 知 $\displaystyle \sum\_\{n=0\}^\infty \frac\{f^\{(n)\}(x\_0)\}\{n!\}(x-x\_0)^n$ 关于 $\displaystyle x\in U(x\_0;\delta)$ 一致收敛. 在 Taylor 展式
\begin\{aligned\} f(x)=\sum\_\{n=1\}^N\frac\{f^\{(n)\}(x\_0)\}\{n!\}(x-x\_0)^n +\frac\{f^\{(N+1)\}(\xi\_N)\}\{(N+1)!\}(x-x\_0)^\{N+1\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
中令 $\displaystyle N\to\infty$ 即得 $\displaystyle f(x)=\sum\_\{n=0\}^\infty \frac\{f^\{(n)\}(x\_0)\}\{n!\}(x-x\_0)^n$.跟锦数学微信公众号. [在线资料](
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发表于 2023-3-5 08:45:03
