## 张祖锦2023年数学专业真题分类70天之第26天
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576、 (6)、 设 $\displaystyle f(x)$ 周期为 $\displaystyle 2\pi$, 在 $\displaystyle [-\pi,\pi)$ 的表达式是 $\displaystyle f(x)=\left\\{\begin\{array\}\{llllllllllll\}-1,&-\pi\leq x < 0,\\\\ 1,&0\leq x < \pi.\end\{array\}\right.$ 则 $\displaystyle f(x)$ 的傅里叶展开式为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (云南大学2023年数学分析考研试题) [Fourier级数 ]
[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
\begin\{aligned\} f(x)\sim \sum\_\{n=1\}^\infty b\_n\sin nx, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中
\begin\{aligned\} n\geq 1\Rightarrow b\_n&=\frac\{2\}\{\pi\}\int\_0^\pi f(x)\sin nx\mathrm\{ d\} x=\left\\{\begin\{array\}\{llllllllllll\}0,&n=2k,\\\\ \frac\{4\}\{(2n-1)\pi\},&n=2k-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\}
故应填 $\displaystyle \frac\{4\}\{\pi\}\sum\_\{k=1\}^\infty \frac\{\sin (2k-1)x\}\{2k-1\}$.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)/
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577、 5、 设 $\displaystyle f(x)$ 是 $\displaystyle \mathbb\{R\}$ 上的周期为 $\displaystyle 2\pi$ 的二阶连续可微函数. 证明: $\displaystyle f(x)$ 的傅里叶级数在 $\displaystyle \mathbb\{R\}$ 上一致收敛于 $\displaystyle f(x)$. (中国科学院大学2023年数学分析考研试题) [Fourier级数 ]
[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 $\displaystyle M=\max\_\{[-\pi,\pi]\}|f''|$, 则当 $\displaystyle n\geq 1$ 时,
\begin\{aligned\} a\_n=&\frac\{1\}\{\pi\}\int\_\{-\pi\}^\pi f(x)\cos nx\mathrm\{ d\} x =\frac\{1\}\{n\pi\}\int\_\{-\pi\}^\pi f(x)\mathrm\{ d\} \sin nx\\\\ \xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\}&-\frac\{1\}\{n\pi\}\int\_\{-\pi\}^\pi f'(x)\sin nx\mathrm\{ d\} x =\frac\{1\}\{n^2\pi\}\int\_\{-\pi\}^\pi f'(x)\mathrm\{ d\} \cos nx\\\\ \xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\}&-\frac\{1\}\{n^2\pi\}\int\_\{-\pi\}^\pi f''(x)\cos nx\mathrm\{ d\} x. \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\} |a\_n|\leq \frac\{1\}\{n^2\pi\} \cdot M \cdot 2\pi=\frac\{2M\}\{n^2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
同理, $\displaystyle |b\_n|\leq \frac\{2M\}\{n^2\}$. 由 Weierstrass 判别法即知
\begin\{aligned\} \frac\{a\_0\}\{2\}+\sum\_\{n=1\}^\infty (a\_n\cos nx+b\_n \sin nx) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
在 $\displaystyle \mathbb\{R\}$ 上一致收敛于 $\displaystyle f(x)$.跟锦数学微信公众号. [在线资料](
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---
578、 (2)、 将 $\displaystyle [0,\pi]$ 上的函数 $\displaystyle f(x)=\frac\{\pi^2\}\{2\}-x^2$ 在 $\displaystyle [-\pi,\pi]$ 上按余弦级数展开, 并求 $\displaystyle \sum\_\{n=1\}^\infty \frac\{(-1)^\{n-1\}\}\{n^2\}$ 的和. (中南大学2023年数学分析考研试题) [Fourier级数 ]
[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
\begin\{aligned\} x^2\sim \frac\{a\_0\}\{2\}+\sum\_\{n=1\}^\infty a\_n\cos nx, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中
\begin\{aligned\} a\_0=&\int\_\{-1\}^1 f(x)\mathrm\{ d\} x=\frac\{2\}\{3\},\\\\ n\geq 1\Rightarrow a\_n=&\int\_\{-1\}^1 f(x)\cos n\pi x\mathrm\{ d\} x=\frac\{4(-1)^n\}\{n^2\pi^2\}. \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\} &x^2\sim \frac\{1\}\{3\}+\sum\_\{n=1\}^\infty \frac\{4(-1)^n\}\{n^2\pi^2\}\cos n\pi x\\\\ \Rightarrow&f(x)\sim \frac\{\pi^2\}\{2\}-\frac\{1\}\{3\}+\sum\_\{n=1\}^\infty \frac\{4(-1)^\{n-1\}\}\{n^2\pi^2\}\cos n\pi x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle x=0$ 得
\begin\{aligned\} \sum\_\{n=1\}^\infty \frac\{(-1)^\{n-1\}\}\{n^2\}=\frac\{1\}\{3\}\cdot\frac\{\pi^2\}\{4\}=\frac\{\pi^2\}\{12\}. \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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---
579、 5、 (15 分) 设 $\displaystyle f(x)$ 是以 $\displaystyle 2\pi$ 为周期的函数, 其傅里叶级数展开式系数为 $\displaystyle a\_0, a\_n, b\_n\ (n=1,2,\cdots)$. 若存在 $\displaystyle L\in (0,1)$, 使得对任意的 $\displaystyle x,y$, 都有
\begin\{aligned\} |f(x)-f(y)|\leq L|x-y|^\alpha, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
求证: $\displaystyle a\_n,b\_n=O\left(\frac\{1\}\{n^\alpha\}\right)$, 即存在常数 $\displaystyle D > 0$, 使得 $\displaystyle |a\_n|, |b\_n|\leq Dn^\alpha$. (中山大学2023年数学分析考研试题) [Fourier级数 ]
[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) / 这就是 Fourier 系数的渐近性质. 由题中不等式知 $\displaystyle f$ 在 $\displaystyle [-\pi,\pi]$ 上一致连续. 从而
\begin\{aligned\} a\_n&=\frac\{1\}\{\pi\}\int\_\{-\pi\}^\pi f(x)\cos nx\mathrm\{ d\} x\qquad(I)\\\\ &\stackrel\{x=t+\frac\{\pi\}\{n\}\}\{=\}\frac\{1\}\{\pi\}\int\_\{-\pi-\frac\{\pi\}\{n\}\}^\{\pi-\frac\{\pi\}\{n\}\} f\left(t+\frac\{\pi\}\{n\}\right)\cos(nt+\pi)\mathrm\{ d\} t\\\\ &=-\frac\{1\}\{\pi\}\int\_\{-\pi\}^\pi f\left(t+\frac\{\pi\}\{n\}\right)\cos nt\mathrm\{ d\} t.\qquad(II) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
最后一步积分限的变换是因为周期函数的缘故. $\displaystyle (I), (II)$ 相加即得
\begin\{aligned\} |a\_n|=&\frac\{1\}\{2\pi\}\left|\int\_\{-\pi\}^\pi \left\[f(x)-f\left(x+\frac\{\pi\}\{n\}\right)\right\] \cos nx\mathrm\{ d\} x\right|\\\\ \leq&\frac\{1\}\{2\pi\}\int\_\{-\pi\}^\pi L \left(\frac\{\pi\}\{n\}\right)^\alpha \mathrm\{ d\} x =L\left(\frac\{\pi\}\{n\}\right)^\alpha=O\left(\frac\{1\}\{n^\alpha\}\right). \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)/
---
580、 (3)、 $\displaystyle \lim\_\{(x,y)\to (0,0)\}\frac\{1-\cos \sqrt\{x^2+y^2\}\}\{\tan(x^2+y^2)\}$. (东北师范大学2023年数学分析考研试题) [多元函数极限与连续 ]
[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) / $\displaystyle \mbox\{原式\}\xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\} \lim\_\{t\to 0\}\frac\{1-\cos t\}\{\tan (t^2)\}\xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\} \frac\{1\}\{2\}$.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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---
581、 (5)、 已知二元函数 $\displaystyle f(x,y)$ 在 $\displaystyle (0,0)$ 处的两个偏导数存在, 则 $\displaystyle f(x,y)$ 在 $\displaystyle (0,0)$ 处连续. (哈尔滨工业大学2023年数学分析考研试题) [多元函数极限与连续 ]
[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) / $\displaystyle \times$. 比如对 $\displaystyle f(x,y)=\left\\{\begin\{array\}\{llllllllllll\}\frac\{xy\}\{x^2+y^2\},&x^2+y^2\neq 0,\\\\ 0,&x^2+y^2=0\end\{array\}\right.$ 而言, $\displaystyle f\_x(0,0)=f\_y(0,0)=0$, 但由
\begin\{aligned\} f(x,kx)=\frac\{k\}\{1+k^2\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle f$ 在原点处不连续.跟锦数学微信公众号. [在线资料](
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---
582、 1、 (10 分) 已知
\begin\{aligned\} f(x,y)=\left\\{\begin\{array\}\{llllllllllll\}x\cos\frac\{1\}\{y\}+y\cos\frac\{1\}\{x\},&xy\neq 0,\\\\ 0,&xy=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\}
判断下列极限是否存在并说明理由:
\begin\{aligned\} \lim\_\{x\to 0\}\lim\_\{y\to 0\}f(x,y), \lim\_\{y\to 0\}\lim\_\{x\to 0\}f(x,y), \lim\_\{x\to 0\atop y\to 0\}f(x,y). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(合肥工业大学2023年数学分析考研试题) [多元函数极限与连续 ]
[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|x\cos\frac\{1\}\{y\}+y\cos\frac\{1\}\{x\}\right|\leq |x|+|y|\xrightarrow\{(x,y)\to (0,0)\}0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \lim\_\{x\to 0\atop y\to 0\}f(x,y)=0$. 又当 $\displaystyle x\neq 0$ 时, 由
\begin\{aligned\} \lim\_\{k\to\infty\}f\left(x,\frac\{1\}\{2k\pi\}\right) =&\lim\_\{k\to\infty\}\left(x+\frac\{1\}\{2k\pi\}\cos\frac\{1\}\{x\}\right)=x,\\\\ \lim\_\{k\to\infty\}f\left(x,\frac\{1\}\{2k\pi+\frac\{\pi\}\{2\}\}\right) =&\lim\_\{k\to\infty\}\frac\{2\}\{(4k+1)\pi\}\cos\frac\{1\}\{x\}=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \lim\_\{y\to 0\}f(x,y)$ 不存在, 进而 $\displaystyle \lim\_\{x\to 0\}\lim\_\{y\to 0\}f(x,y)$ 不存在. 同理, $\displaystyle \lim\_\{y\to 0\}\lim\_\{x\to 0\}f(x,y)$ 不存在.跟锦数学微信公众号. [在线资料](
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---
583、 11、 设一元函数 $\displaystyle f(x)$ 在 $\displaystyle [0,+\infty)$ 上可导, 且存在两个正数 $\displaystyle A < B$ 满足 $\displaystyle A < |f'(x)| < B$. 证明: $\displaystyle f(\sqrt\{x^2+y^2\})$ 在 $\displaystyle \mathbb\{R\}^2$ 上一致连续, 但 $\displaystyle f(x^2+y^2)$ 在 $\displaystyle \mathbb\{R\}^2$ 上不一致连续. (华东师范大学2023年数学分析考研试题) [多元函数极限与连续 ]
[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)、 设 $\displaystyle z\_i=x\_i+\mathrm\{ i\} y\_i, i=1,2$, 则
\begin\{aligned\} &\left|\sqrt\{x\_1^2+y\_1^2\}-\sqrt\{x\_2^2+y\_2^2\}\right| =\left||Oz\_1|-|Oz\_2|\right|\\\\ \leq& |z\_1-z\_2| =\sqrt\{(x\_1-x\_2)^2+(y\_1-y\_2)^2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 对 $\displaystyle \forall\ \varepsilon > 0,\exists\ \delta=\frac\{\varepsilon\}\{M\} > 0, \forall\ (x\_1,y\_1),(x\_2,y\_2)\in\mathbb\{R\}^2$:
\begin\{aligned\} &\sqrt\{(x\_1-x\_2)^2+(y\_1-y\_2)^2\} < \delta\\\\ \Rightarrow&|f(\sqrt\{x\_1^2+y\_1^2\})-f(\sqrt\{x\_2^2+y\_2^2\})|\\\\ &\xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\} |f'(\xi)|\left|\sqrt\{x\_1^2+y\_1^2\}-\sqrt\{x\_2^2+y\_2^2\}\right| < B\delta=\varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle f(\sqrt\{x^2+y^2\})$ 在 $\displaystyle \mathbb\{R\}^2$ 上一致连续. (3)、 取
\begin\{aligned\} x\_n=n, y\_n=0;\quad u\_n=n+\frac\{1\}\{n\}, v\_n=0, \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\} (x\_n-u\_n)^2+(y\_n-v\_n)^2=\frac\{1\}\{n^2\}\xrightarrow\{n\to\infty\}0, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
但
\begin\{aligned\} |f\left(x\_n^2+y\_n^2\right)-f\left(u\_n^2+v\_n^2\right)| =&\left|f(n^2)-f\left(n^2+2+\frac\{1\}\{n^2\}\right)\right|\\\\ \xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\}& |f'(\xi\_n)|\left(2+\frac\{1\}\{n^2\}\right)\geq 2A. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle f(x^2+y^2)$ 在 $\displaystyle \mathbb\{R\}^2$ 上不一致连续.跟锦数学微信公众号. [在线资料](
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---
584、 7、 $\displaystyle f(x,y)$ 在开域 $\displaystyle \varOmega$ 上分别对变量 $\displaystyle x,y$ 是连续的, 并且每当固定 $\displaystyle x$ 时, $\displaystyle f(x,y)$ 对 $\displaystyle y$ 单调下降. 证明: $\displaystyle f(x,y)$ 在开域 $\displaystyle \varOmega$ 上连续. (华南理工大学2023年数学分析考研试题) [多元函数极限与连续 ]
[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) / 通过调换 $\displaystyle x,y$ 后不妨设 $\displaystyle f$ 关于 $\displaystyle x$ 递减; 再通过 $\displaystyle f\mapsto -f$ 不妨设 $\displaystyle f$ 关于 $\displaystyle x$ 递增. 对任一固定的 $\displaystyle (x\_0,y\_0)\in \varOmega$ 及 $\displaystyle \varOmega$ 是区域知
\begin\{aligned\} \exists\ \tilde\{\\}delta > 0,\mathrm\{ s.t.\} [x\_0-\tilde\{\\}delta,x\_0+\tilde\{\\}delta]\times [y\_0-\tilde\{\\}delta,y\_0+\tilde\{\\}delta]\subset \varOmega. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle f$ 关于 $\displaystyle x$ 连续知对任一固定的 $\displaystyle \varepsilon > 0$,
\begin\{aligned\} \exists\ \delta\_1\in (0,\tilde\{\\}delta),\mathrm\{ s.t.\} \left\\{\begin\{array\}\{l\} |f(x\_0-\delta,y\_0)-f(x\_0,y\_0)| < \frac\{\varepsilon\}\{2\},\\\\ |f(x\_0+\delta,y\_0)-f(x\_0,y\_0)| < \frac\{\varepsilon\}\{2\}. \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 $\displaystyle f$ 关于 $\displaystyle y$ 连续知
\begin\{aligned\} \exists\ \delta\_2\in (0,\tilde\{\\}delta),\mathrm\{ s.t.\} |y-y\_0|\leq \delta\_2\Rightarrow \left\\{\begin\{array\}\{l\} |f(x\_0-\delta\_1,y)-f(x\_0-\delta\_1,y\_0)| < \frac\{\varepsilon\}\{2\},\\\\ |f(x\_0+\delta\_1,y)-f(x\_0+\delta\_1,y\_0)| < \frac\{\varepsilon\}\{2\}. \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\}
最后由 $\displaystyle f$ 对 $\displaystyle x$ 单增知当
\begin\{aligned\} |x-x\_0| < \delta\_1,\quad |y-y\_0| < \delta\_2 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
时,
\begin\{aligned\} \begin\{array\}\{rlll\} f(x,y)\leq& f(x\_0+\delta\_1,y)&\quad f(x,y)&\geq f(x\_0-\delta,y)\\\\ < & f(x\_0+\delta\_1,y\_0)+\frac\{\varepsilon\}\{2\}&\quad & > f(x\_0-\delta,y\_0)-\frac\{\varepsilon\}\{2\}\\\\ < & f(x\_0,y\_0)+\varepsilon,&\quad &\geq f(x\_0,y\_0)-\varepsilon. \end\{array\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
此即
\begin\{aligned\} |f(x,y)-f(x\_0,y\_0)| < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
585、 (4)、 $\displaystyle \lim\_\{(x,y)\to (0,0)\}\frac\{x^2+y^2\}\{3-\sqrt\{(3+x^2)(3+y^2)\}\}$. (华南师范大学2023年数学分析考研试题) [多元函数极限与连续 ]
[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) /
\begin\{aligned\} \mbox\{原式\}=&\lim\_\{(x,y)\to (0,0)\}\frac\{x^2+y^2\}\{3-\sqrt\{(3+x^2)(3+y^2)\}\}\\\\ =&\lim\_\{(x,y)\to (0,0)\}\frac\{x^2+y^2\}\{9-(3+x^2)(3+y^2)\}\left\[3+\sqrt\{(3+x^2)(3+y^2)\}\right\]\\\\ =&-6\lim\_\{(x,y)\to (0,0)\}\frac\{x^2+y^2\}\{3(x^2+y^2)+x^2y^2\} =-6\cdot\frac\{1\}\{3\}=-2. \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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---
586、 (0-17)、 求极限 $\displaystyle \lim\_\{(x,y)\to (0,2)\} (1+y\sin^2x)^\frac\{1\}\{x\ln (1+2x)\}$. (吉林师范大学2023年(学科数学)数学分析考研试题) [多元函数极限与连续 ]
[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) /
\begin\{aligned\} \mbox\{原式\}=&\mathrm\{exp\}\left\[\lim\_\{(x,y)\to (0,2)\} \frac\{\ln(1+y\sin^2x)\}\{x\ln(1+2x)\}\right\]\\\\ \xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\}&\mathrm\{exp\}\left\[\lim\_\{(x,y)\to (0,2)\} \frac\{y\sin^2x\}\{x\cdot 2x\}\right\] =\mathrm\{exp\}\left\[\frac\{1\}\{2\}\lim\_\{x\to 0\}\frac\{\sin^2x\}\{x^2\}\cdot \lim\_\{y\to 2\}y\right\] =\mathrm\{e\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
587、 6、 (15 分) $\displaystyle f(x,y)=\left\\{\begin\{array\}\{llllllllllll\}x\sin\frac\{1\}\{y\}+y\sin\frac\{1\}\{x\},&xy\neq 0,\\\\ 0,&xy=0.\end\{array\}\right.$ 讨论下面三个极限:
\begin\{aligned\} \lim\_\{(x,y)\to (0,0)\}f(x,y), \lim\_\{x\to 0\}\lim\_\{y\to 0\}f(x,y), \lim\_\{y\to 0\}\lim\_\{x\to 0\}f(x,y). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(山西大学2023年数学分析考研试题) [多元函数极限与连续 ]
[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 $\displaystyle \left|x\sin\frac\{1\}\{y\}+y\sin\frac\{1\}\{x\}\right|\leq |x|+|y|$ 即知 $\displaystyle \vlmc\{(x,y)\}\{(0,0)\}f(x,y)=0$. (2) 当 $\displaystyle x\neq 0$ 时, $\displaystyle \vlmc\{y\}\{0\} x\sin\frac\{1\}\{y\}$ 不存在, 而 $\displaystyle \vlmc\{x\}\{0\}\vlmc\{y\}\{0\}f(x,y)$ 不存在. (3) 当 $\displaystyle y\neq 0$ 时, $\displaystyle \vlmc\{x\}\{0\} y\sin\frac\{1\}\{x\}$ 不存在, 而 $\displaystyle \vlmc\{y\}\{0\}\vlmc\{x\}\{0\}f(x,y)$ 不存在.跟锦数学微信公众号. [在线资料](
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---
588、 7、 (本题 10 分) $\displaystyle f(x,y)$ 在 $\displaystyle [0,1]\times [0,1]$ 上连续,
\begin\{aligned\} g(x)=\sup\_\{y\in [0,1]\}f(x,y), x\in [0,1]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
证明: $\displaystyle g(x)$ 在 $\displaystyle [0,1]$ 上连续. (四川大学2023年数学分析考研试题) [多元函数极限与连续 ]
[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 $\displaystyle f(x,y)$ 在 $\displaystyle [0,1]\times [0,1]$ 上连续知其一致连续:
\begin\{aligned\} &\forall\ \varepsilon > 0,\exists\ \delta > 0,\mathrm\{ s.t.\} \forall\ (x',y'), (x,y)\in [0,1]^2,\\\\ &|x'-x| < \delta, |y'-y| < \delta, |f(x',y')-f(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\}
特别地, 当 $\displaystyle |x'-x| < \delta$ 时,
\begin\{aligned\} &|f(x',y)-f(x,y)| < \varepsilon, \forall\ y\in[0,1]\\\\ \Rightarrow&f(x,y)-\varepsilon < f(x',y) < f(x,y)+\varepsilon, \forall\ y\in [0,1]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
对 $\displaystyle y\in [0,1]$ 取上确界知
\begin\{aligned\} g(x)-\varepsilon\leq g(x')\leq g(x)+\varepsilon\Rightarrow |g(x')-g(x)|\leq\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 $\displaystyle g$ 在 $\displaystyle [0,1]$ 上一致连续, 而连续.跟锦数学微信公众号. [在线资料](
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---
589、 5、 设 $\displaystyle f$ 是 $\displaystyle \mathbb\{R\}^n$ 上的函数. 求证: $\displaystyle f$ 连续的充要条件是开集的原像是开集. (同济大学2023年数学分析考研试题) [多元函数极限与连续 ]
[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)、 $\displaystyle \Rightarrow$: 对 $\displaystyle \mathbb\{R\}$ 中的开集 $\displaystyle O$, 往证 $\displaystyle f^\{-1\}(O)$ 是开集. 对 $\displaystyle \forall\ x\_0\in f^\{-1\}(O)$, 我们有 $\displaystyle f(x\_0)\in O$, 而
\begin\{aligned\} \exists\ \varepsilon > 0,\mathrm\{ s.t.\} U\left(f(x\_0);\varepsilon\right)\subset O. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
再由 $\displaystyle f$ 在 $\displaystyle x\_0$ 处连续知
\begin\{aligned\} \exists\ \delta > 0,\mathrm\{ s.t.\} \forall\ x\in U(x\_0;\delta), |f(x)-f(x\_0)| < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
此即
\begin\{aligned\} f\left(U(x\_0;\delta)\right)\subset U\left(f(x\_0);\varepsilon\right)\subset O \Rightarrow U(x\_0;\delta)\subset f^\{-1\}(O). \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 $\displaystyle f^\{-1\}(O)$ 是开集. (2)、 $\displaystyle \Leftarrow$: 对 $\displaystyle \forall\ x\_0\in\mathbb\{R\}$, $\displaystyle \forall\ \varepsilon > 0$, $\displaystyle U\left(f(x\_0);\varepsilon\right)$ 是包含 $\displaystyle f(x\_0)$ 的开集. 由充分性假设知 $\displaystyle f^\{-1\}\left(U\left(f(x\_0);\varepsilon\right)\right)$ 是包含 $\displaystyle x\_0$ 的开集. 从而
\begin\{aligned\} \exists\ \delta > 0,\mathrm\{ s.t.\} &U(x\_0;\delta)\subset f^\{-1\}\left(U\left(f(x\_0);\varepsilon\right)\right)\\\\ \Rightarrow& f\left(U(x\_0;\delta)\right)\subset U\left(f(x\_0);\varepsilon\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
此即 $\displaystyle \forall\ x\in U(x\_0;\delta), |f(x)-f(x\_0)| < \varepsilon$. 这就证明了 $\displaystyle f$ 在 $\displaystyle x\_0$ 连续. 由 $\displaystyle x\_0$ 的任意性知 $\displaystyle f$ 在 $\displaystyle \mathbb\{R\}$ 上连续.跟锦数学微信公众号. [在线资料](
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---
590、 (2)、 对点列 $\displaystyle z\_n=(x\_n,y\_n)\ (n=1,2,\cdots)$, 有
\begin\{aligned\} \varliminf\_\{n\to\infty\}\left\Vert z\_n\right\Vert =\alpha, \varlimsup\_\{n\to\infty\}\left\Vert z\_n\right\Vert =\beta, \lim\_\{n\to\infty\}\left\Vert z\_\{n+1\}-z\_n\right\Vert =0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
证明: 对任意的 $\displaystyle \gamma\in (\alpha,\beta), x^2+y^2=\gamma^2$ 上至少存在 $\displaystyle \left\\{z\_n\right\\}$ 的一个聚点. (武汉大学2023年数学分析考研试题) [多元函数极限与连续 ]
[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) / (2-1)、 先证明一个结论. 设有界数列 $\displaystyle \left\\{t\_n\right\\}$ 满足
\begin\{aligned\} \varliminf\_\{n\to\infty\}t\_n=A < B=\varlimsup\_\{n\to\infty\}t\_n, \lim\_\{n\to\infty\}(t\_\{n+1\}-t\_n)=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle \left\\{t\_n\right\\}$ 的聚点全体恰好构成 $\displaystyle [A,B]$. 事实上, 设 $\displaystyle \left\\{t\_n\right\\}$ 的聚点全体为 $\displaystyle M$, 则由上下极限的定义与性质知 $\displaystyle M\subset [A,B]$. 而仅需证明 $\displaystyle (A,B)\subset M$. 用反证法. 若 $\displaystyle (A,B)\not\subset M$, 则 $\displaystyle \exists\ A < c < B,\mathrm\{ s.t.\} c\not\in M.$ 我们断言: $\displaystyle \exists\ 0 < \delta < \min\left\\{c-A,B-c\right\\}$, 使得 $\displaystyle (c-\delta,c+\delta)$ 中仅含有 $\displaystyle \left\\{t\_n\right\\}$ 中有限多项. 事实上, 若 $\displaystyle \forall\ \delta > 0, (c-\delta,c+\delta)$ 都含有 $\displaystyle \left\\{t\_n\right\\}$ 中无限多项, 而 $\displaystyle c$ 为 $\displaystyle \left\\{t\_n\right\\}$ 的聚点, $\displaystyle c\in M$, 这是一个矛盾. 这样, \begin\{aligned\} \exists\ N,\mathrm\{ s.t.\} n\geq N\Rightarrow t\_n\leq c-\delta\mbox\{或 \}t\_n\geq c+\delta. \qquad(1.6.10: eq1)\tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
又由 $\displaystyle \lim\_\{n\to\infty\}(t\_\{n+1\}-t\_n)=0$ 知 \begin\{aligned\} \exists\ N'\geq N,\mathrm\{ s.t.\} n\geq N'\Rightarrow |t\_\{n+1\}-t\_n| < 2\delta. \qquad(1.6.10: eq2)\tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 (1.6.10: eq1), 不妨设 $\displaystyle t\_\{N'\}\leq c-\delta$. 又由 (1.6.10: eq1)-(1.6.10: eq2) 知 $\displaystyle t\_\{N'+1\}\leq c-\delta$. 如此一直做下去, 我们得到 $\displaystyle n\geq N'\Rightarrow t\_n\leq c-\delta. $ 令 $\displaystyle n\to\infty$, 得 $\displaystyle B=\varlimsup\_\{n\to\infty\}t\_n\leq c-\delta < B .$ 这是一个矛盾. 故有结论. (2-2)、 回到题目. 设 $\displaystyle t\_n=\left\Vert z\_n\right\Vert $, 则由三角不等式及题设知 $\displaystyle \lim\_\{n\to\infty\}(t\_\{n+1\}-t\_n)=0$. 对 $\displaystyle \forall\ \gamma\in (\alpha,\beta)$, 由第 i 步知
\begin\{aligned\} \exists\ n\_k,\mathrm\{ s.t.\} \gamma=\lim\_\{k\to\infty\}t\_\{n\_k\}=\lim\_\{k\to\infty\}\left\Vert z\_\{n\_k\}\right\Vert . \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
再由致密性定理, $\displaystyle \exists\ n\_\{k\_i\}, \left\Vert z\_0\right\Vert \leq \beta,\mathrm\{ s.t.\} \lim\_\{i\to\infty\}z\_\{n\_\{k\_i\}\}=z\_0$,
\begin\{aligned\} \gamma=\lim\_\{i\to\infty\}\left\Vert z\_\{n\_\{k\_i\}\}\right\Vert =\left\Vert z\_0\right\Vert . \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
如此, $\displaystyle z\_0$ 就是 $\displaystyle x^2+y^2=\gamma^2$ 上的点, 且是 $\displaystyle \left\\{z\_n\right\\}$ 的一个聚点.跟锦数学微信公众号. [在线资料](
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---
591、 (7)、 $\displaystyle E=\left\\{(x,y)\in\mathbb\{R\}^2; x,y\in\mathbb\{Q\}\right\\}$, 则 $\displaystyle \partial E=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (西安交通大学2023年数学分析考研试题) [多元函数极限与连续 ]
[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) / $\displaystyle \mathbb\{R\}^2$ 上任一点的任一邻域都有有理点, 也有无理点, 从而 $\displaystyle \partial E=\mathbb\{R\}^2$.跟锦数学微信公众号. [在线资料](
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---
592、 6、 (20 分) 证明: $\displaystyle \lim\_\{(x,y)\to (0,0)\}x^y$ 不存在, 其中 $\displaystyle x > 0, y > 0$. (西南财经大学2023年数学分析考研试题) [多元函数极限与连续 ]
[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 $\displaystyle f(x,y)=x^y$, 则
\begin\{aligned\} \lim\_\{n\to\infty\} f\left(\frac\{1\}\{n\},\frac\{1\}\{n\}\right) =&\lim\_\{n\to\infty\} \left(\frac\{1\}\{n\}\right)^\frac\{1\}\{n\} =\mathrm\{exp\}\left\[-\lim\_\{n\to\infty\}\frac\{1\}\{n\}\ln n\right\]\\\\ \xlongequal\{\tiny\mbox\{Stolz\}\}&\mathrm\{exp\}\left\[-\lim\_\{n\to\infty\}\ln \left(1+\frac\{1\}\{n\}\right)\right\]=1,\\\\ \lim\_\{n\to\infty\} f\left(\frac\{1\}\{n\}, \ln n\right) =&\lim\_\{n\to\infty\}\left(\frac\{1\}\{n\}\right)^\{\ln n\} =\mathrm\{exp\}\left\[-\lim\_\{n\to\infty\}\ln^2n\right\]=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \lim\_\{(x,y)\to (0,0)\}x^y$ 不存在.跟锦数学微信公众号. [在线资料](
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---
593、 (2)、 $\displaystyle \lim\_\{x\to 0\atop y\to 2\}\frac\{\sin xy\}\{x\}=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (重庆师范大学2023年数学分析考研试题) [多元函数极限与连续 ]
[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) /
\begin\{aligned\} \mbox\{原式\}=&\lim\_\{x\to 0\atop y\to 2\}\frac\{\sin xy\}\{xy\}\cdot y =\lim\_\{t\to 0\}\frac\{\sin t\}\{t\}\cdot \lim\_\{y\to 2\}y=1\cdot 2=2. \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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---
594、 7、 (20 分) 设二元函数
\begin\{aligned\} f(x,y)=\left\\{\begin\{array\}\{llllllllllll\}\frac\{2xy^3\}\{x^2+y^4\},&x^2+y^2\neq 0,\\\\ 0,&x^2+y^2=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\}
讨论 $\displaystyle f$ 在原点的连续性, 偏导数的存在性及 $\displaystyle f$ 在原点的可微性. (安徽大学2023年高等代数考研试题) [多元函数微分学 ]
[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\} 0\leq \left|\frac\{2xy^3\}\{x^2+y^4\}\right| \leq \frac\{2|xy^2|\}\{x^2+y^4\}|y|\leq |y|\xrightarrow\{y\to 0\}0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle f$ 在原点连续. 易知 $\displaystyle f\_x(0,0)=f\_y(0,0)=0$. 最后由
\begin\{aligned\} &\left|\frac\{f(x,y)-f(0,0)-f\_x(0,0)x-f\_y(0,0)y\}\{\sqrt\{x^2+y^2\}\}\right|\\\\ =&\left|\frac\{2xy^3\}\{(x^2+y^4)\sqrt\{x^2+y^2\}\}\right| \stackrel\{x=ky^2\}\{=\}\frac\{2k\}\{(k^2+1)\sqrt\{k^2y^2+1\}\}\xrightarrow\{y\to 0\}\frac\{2k\}\{k^2+1\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle f$ 在原点不可微.跟锦数学微信公众号. [在线资料](
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595、 8、 (20 分) 证明方程 $\displaystyle x+\frac\{1\}\{2\}y^2+\frac\{1\}\{2\}z+\sin z=0$ 在 $\displaystyle (0,0,0)$ 附近唯一确定了隐函数 $\displaystyle z=f(x,y)$, 进一步将 $\displaystyle f(x,y)$ 在 $\displaystyle (0,0)$ 处展开为带佩亚诺型余项的泰勒公式, 展开到二阶. (安徽大学2023年高等代数考研试题) [多元函数微分学 ]
[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 $\displaystyle F=x+\frac\{1\}\{2\}y^2+\frac\{1\}\{2\}z+\sin z$, 则
\begin\{aligned\} F\_z(0,0,0)=\left.\left(\frac\{1\}\{2\}+\cos z\right)\right||\_\{z=0\}=\frac\{3\}\{2\}\neq 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由隐函数存在定理知 $\displaystyle F=0$ 在 $\displaystyle (0,0,0)$ 附近唯一确定了隐函数 $\displaystyle z=f(x,y)$. 再者, 对由 $\displaystyle F(x,y,z)=0$ 所确定的隐函数 $\displaystyle z=z(x,y)$ 有 $\displaystyle F\_x'+F\_z'z\_x'=0, F\_y'+F\_z'z\_y'=0;$
\begin\{aligned\} F\_\{xx\}''+F\_\{xz\}''z\_x'+(F\_\{zx\}''+F\_\{zz\}''z\_x')z\_x'+F\_z'z\_\{xx\}''&=0,\\\\ F\_\{xy\}''+F\_\{xz\}''z\_y'+(F\_\{zy\}''+F\_\{zz\}''z\_y')z\_x'+F\_z'z\_\{xy\}''&=0,\\\\ F\_\{yy\}''+F\_\{yz\}''z\_y'+(F\_\{zy\}''+F\_\{zz\}''z\_y')z\_y'+F\_z'z\_\{yy\}''&=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是 $\displaystyle \displaystyle z\_x'=-\frac\{F\_x'\}\{F\_z'\},\quad z\_y'=-\frac\{F\_y'\}\{F\_z'\};$
\begin\{aligned\} z\_\{xx\}''&=-\frac\{z\_x'^2F\_\{zz\}''+2z\_x'F\_\{xz\}''+F\_\{xx\}''\}\{F\_z'\} =-\frac\{F\_x'^2F\_\{zz\}''-2F\_x'F\_z'F\_\{xz\}''+F\_z'^2F\_\{xx\}''\}\{F\_z'^3\},\\\\ z\_\{xy\}''&=-\frac\{z\_x'z\_y'F\_\{zz\}''+z\_x'F\_\{yz\}''+z\_y'F\_\{xz\}''+F\_\{xy\}''\}\{F\_z'\}\\\\ &=-\frac\{F\_x'F\_y'F\_\{zz\}''+F\_x'F\_z'F\_\{yz\}''+F\_y'F\_z'F\_\{xz\}''+F\_z'^2F\_\{xy\}''\}\{F\_y'^3\},\\\\ z\_\{yy\}'&=-\frac\{z\_y'^2F\_\{zz\}''+2z\_y'F\_\{yz\}''+F\_\{yy\}''\}\{F\_z'\} =-\frac\{F\_y'^2F\_\{zz\}''-2F\_y'F\_z'F\_\{yz\}''+F\_z'^2F\_\{yy\}''\}\{F\_z'^3\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
将 $\displaystyle F(x,y,z)$ 代入, 即知在 $\displaystyle x=0, y=0, z=0$ 处, $\displaystyle z\_x'=-\frac\{2\}\{3\}, z\_y'=0, z\_\{xx\}''=0, z\_\{xy\}''=0, z\_\{yy\}''=-\frac\{2\}\{3\}.$ 故 $\displaystyle f(x,y)$ 在 $\displaystyle (0,0)$ 处展开为带佩亚诺型余项的泰勒公式为
\begin\{aligned\} z=&f(x,y)=-\frac\{2\}\{3\}x+\frac\{1\}\{2\}\left(-\frac\{2\}\{3\}\right)y^2+o\left((x^2+y^2)\right)\\\\ =&-\frac\{2\}\{3\}x-\frac\{1\}\{3\}y^2+o\left((x^2+y^2)\right). \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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596、 (2)、 已知函数 $\displaystyle u(x,y)$ 满足
\begin\{aligned\} 2\frac\{\partial^2u\}\{\partial x^2\}+3\frac\{\partial u\}\{\partial x\}+3\frac\{\partial u\}\{\partial y\}-2\frac\{\partial^2u\}\{\partial y^2\}=0, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
求 $\displaystyle a,b$ 的值, 使得在变换 $\displaystyle u(x,y)=v(x,y)\mathrm\{e\}^\{ax+by\}$ 下上述等式可以化为函数 $\displaystyle v(x,y)$ 的不含一阶偏导数的等式. (北京科技大学2023年数学分析考研试题) [多元函数微分学 ]
[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 $\displaystyle u\_x=v\_x\mathrm\{e\}^\{ax+by\}+av\mathrm\{e\}^\{ax+by\}, u\_y=v\_y\mathrm\{e\}^\{ax+by\}+bv\mathrm\{e\}^\{ax+by\}$ 知
\begin\{aligned\} u\_\{xx\}=&v\_\{xx\}\mathrm\{e\}^\{ax+by\}+2av\_x\mathrm\{e\}^\{ax+by\}+a^2v\mathrm\{e\}^\{ax+by\},\\\\ u\_\{yy\}=&v\_\{yy\}\mathrm\{e\}^\{ax+by\}+2bv\_y\mathrm\{e\}^\{ax+by\}+b^2v\mathrm\{e\}^\{ax+by\}. \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\} 0=&2u\_\{xx\}+3u\_x+3u\_y-2u\_\{yy\}\\\\ =&2(v\_\{xx\}-v\_\{yy\})\mathrm\{e\}^\{ax+by\}+(4a+3)v\_x\mathrm\{e\}^\{ax+by\}\\\\ &+(-4b+3)v\_y\mathrm\{e\}^\{ax+by\}+(a^2-b^2+3a+3b)v\mathrm\{e\}^\{ax+by\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是当且仅当 $\displaystyle a=-\frac\{3\}\{4\}, b=\frac\{3\}\{4\}$ 时, 题中等式可以化为函数 $\displaystyle v(x,y)$ 的不含一阶偏导数的等式 $\displaystyle v\_\{xx\}-v\_\{yy\}=0$.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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597、 5、 (15 分) 已知 $\displaystyle f(x,y)=|x-y|\varphi(x,y)$, 其中 $\displaystyle \varphi$ 在原点处连续. 证明: $\displaystyle \varphi(0,0)=0$ 当且仅当 $\displaystyle f(x,y)$ 在 $\displaystyle (0,0)$ 处可微. (北京师范大学2023年数学分析考研试题) [多元函数微分学 ]
[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)、 $\displaystyle \Rightarrow$:
\begin\{aligned\} f\_x(0,0)=\lim\_\{x\to 0\}\frac\{f(x,0)-f(0,0)\}\{x\}=\lim\_\{x\to 0\}\frac\{|x|\varphi(x,0)\}\{x\} \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\}
同理, $\displaystyle f\_y(0,0)=0$. 故由
\begin\{aligned\} &\left|\frac\{f(x,y)-f(0,0)-f\_x(0,0)x-f\_y(0,0)y\}\{\sqrt\{x^2+y^2\}\}\right|\\\\ =&\frac\{|x-y|\cdot |\varphi(x,y)|\}\{\sqrt\{x^2+y^2\}\} \leq 2|\varphi(x,y)|\xrightarrow\{(x,y)\to(0,0)\}0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle f$ 在 $\displaystyle (0,0)$ 处可微. (2)、 $\displaystyle \Leftarrow$: 由 $\displaystyle f$ 在原点可微知
\begin\{aligned\} f\_x(0,0)=&\lim\_\{x\to 0\}\frac\{f(x,0)-f(0,0)\}\{x\}=\lim\_\{x\to 0\}\frac\{|x|\varphi(x,0)\}\{x\}\\\\ =&\varphi(0,0)\lim\_\{x\to 0\}\frac\{|x|\}\{x\} =\left\\{\begin\{array\}\{llllllllllll\}\varphi(0,0)\lim\_\{x\to 0^+\}1=\varphi(0,0),\\\\ \varphi(0,0)\lim\_\{x\to 0^-\}(-1)=-\varphi(0,0).\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\}
存在. 故 $\displaystyle 2\varphi(0,0)=0\Rightarrow \varphi(0,0)=0$.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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598、 6、 (15 分) 求抛物面 $\displaystyle x^2+y^2=2az$ 被 $\displaystyle x^2+xy+y^2=a^2$ ($a > 0$) 所截曲线的高度 ($z$ 的坐标) 的最值. (北京师范大学2023年数学分析考研试题) [多元函数微分学 ]
[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
\begin\{aligned\} L=x^2+y^2+\lambda(x^2+xy+y^2-a^2), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则由
\begin\{aligned\} L\_x&=2x+2\lambda x+\lambda y=0,\\\\ L\_y&=2y+\lambda x+2\lambda y=0,\\\\ L\_\lambda&=x^2+xy+y^2-a^2=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \lambda\neq 0$. 从 $\displaystyle L\_x=0$ 求出 $\displaystyle y$, 代入 $\displaystyle L\_y=0$ 后即知 $\displaystyle \lambda=-2\mbox\{或\} -\frac\{2\}\{3\}$. 再代入 $\displaystyle L\_\lambda=0$ 求出
\begin\{aligned\} (x,y)\in \left\\{(-a,a),(a,-a), \left(-\frac\{a\}\{\sqrt\{3\}\},-\frac\{a\}\{\sqrt\{3\}\}\right),\left(\frac\{a\}\{\sqrt\{3\}\},\frac\{a\}\{\sqrt\{3\}\}\right)\right\\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
算得 $\displaystyle z=\frac\{x^2+y^2\}\{2a\}$ 在上述四点的值分别为 $\displaystyle a,a,\frac\{a\}\{3\},\frac\{a\}\{3\}$, 而 $\displaystyle \max z=a$.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)/
发表于 2023-3-5 09:09:53
