## 张祖锦2023年数学专业真题分类70天之第05天
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93、 1、 (5 分) 求数列极限 $\displaystyle \lim\_\{n\to\infty\}(1+2^n+3^n)^\frac\{1\}\{n\}$. (中国矿业大学(北京)2023年数学分析考研试题) [数列极限 ]
[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 $\displaystyle 3\leq (1+2^n+3^n)^\frac\{1\}\{n\}\leq 3^\frac\{1\}\{n\}\cdot 3$ 及迫敛性知原式 $\displaystyle =3$.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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---
94、 1、 (15 分) 求极限
\begin\{aligned\} \lim\_\{n\to\infty\}\left(\frac\{\sqrt[n]\{a\}+\sqrt[n]\{b\}+\sqrt[n]\{c\}\}\{3\}\right)^n, a > 0, b > 0, c > 0. \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) [数列极限 ]
[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\[\lim\_\{n\to\infty\}n\ln \frac\{a^\frac\{1\}\{n\}+b^\frac\{1\}\{n\}+c^\frac\{1\}\{n\}\}\{3\}\right\]\\\\ \xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\}& \mathrm\{exp\}\left\[\lim\_\{n\to\infty\}n \left(\frac\{a^\frac\{1\}\{n\}+b^\frac\{1\}\{n\}+c^\frac\{1\}\{n\}\}\{3\}-1\right)\right\]\\\\ \xlongequal[\tiny\mbox\{原理\}]\{\tiny\mbox\{归结\}\}&\mathrm\{exp\}\left\[\lim\_\{x\to 0\}\frac\{a^x+b^x+c^x-3\}\{3x\}\right\]\\\\ \xlongequal\{\tiny\mbox\{L'Hospital\}\}& \mathrm\{exp\}\left\[\lim\_\{x\to 0\}\frac\{a^x\ln a+b^x\ln b+c^x\ln c\}\{3\}\right\]\\\\ =&\mathrm\{exp\}\left\[\frac\{\ln a+\ln b+\ln c\}\{3\}\right\]=\sqrt[3]\{abc\}. \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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95、 1、 填空题. (1)、 $\displaystyle \lim\_\{n\to\infty\}\left(\frac\{1\}\{n+\sqrt\{1\}\}+\frac\{1\}\{n+\sqrt\{2\}\}+\cdots+\frac\{1\}\{n+\sqrt\{n\}\}\right)=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (重庆师范大学2023年数学分析考研试题) [数列极限 ]
[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\} 1=\sum\_\{k=1\}^n \frac\{1\}\{n\} > \sum\_\{k=1\}^n \frac\{1\}\{n+\sqrt\{k\}\} > \frac\{1\}\{n+\sqrt\{n\}\}\sum\_\{k=1\}^n 1=\frac\{n\}\{n+\sqrt\{n\}\} =\frac\{1\}\{1+\frac\{1\}\{\sqrt\{n\}\}\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及迫敛性知原式 $\displaystyle =1$.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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96、 (2)、 $\displaystyle \lim\_\{x\to 0\}\frac\{\displaystyle x-\int\_0^x (1+\sin^2t)^2\mathrm\{ d\} t\}\{x^2\sin x\}$. (安徽大学2023年数学分析考研试题) [函数极限 ]
[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\}\xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\}&\lim\_\{x\to 0\}\frac\{\displaystyle x-\int\_0^x (1+\sin^2t)^2\mathrm\{ d\} t\}\{x^3\} \xlongequal\{\tiny\mbox\{L'Hospital\}\} \lim\_\{x\to 0\}\frac\{1-(1+\sin^2x)^2\}\{3x^2\}\\\\ =&\frac\{1\}\{3\}\lim\_\{x\to 0\}\frac\{[1+(1+\sin^2x)][1-(1+\sin^2x)]\}\{x^2\} =\frac\{1\}\{3\}\cdot 2\cdot (-1)=-\frac\{2\}\{3\}. \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)/
---
97、 1、 计算题 (每题 10 分, 共 40 分). (1)、 求极限 $\displaystyle \lim\_\{x\to\infty\}\left(\cos\frac\{1\}\{x\}\right)^\{x^2\}$. (北京科技大学2023年数学分析考研试题) [函数极限 ]
[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\[\lim\_\{t\to 0\}\frac\{1\}\{t^2\}\ln \cos t\right\] \xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\} \mathrm\{exp\}\left\[\lim\_\{t\to 0\}\frac\{1\}\{t^2\}(\cos t-1)\right\] \xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\} \mathrm\{e\}^\{-\frac\{1\}\{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](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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98、 1、 已知 $\displaystyle \lim\_\{x\to 0\}\frac\{\cos 2x-\sqrt\{\cos 2x\}\}\{x^k\}=a\ (a\neq 0)$, 求 $\displaystyle a$ 和 $\displaystyle k$ 的值. (北京邮电大学2023年数学分析考研试题) [函数极限 ]
[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\} \cos 2x-\sqrt\{\cos 2x\}=&\frac\{\cos^22x-\cos 2x\}\{\cos2x+\sqrt\{\cos 2x\}\} \sim \frac\{\cos 2x(\cos 2x-1)\}\{2\}\\\\ \sim&\frac\{1\}\{2\} \left\[-\frac\{(2x)^2\}\{2\}\right\] =-x^2, x\to 0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle k=2, a=-1$.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](
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---
99、 (3)、 计算极限
\begin\{aligned\} \lim\_\{x\to 1\}\left(\frac\{m\}\{1-x^m\}-\frac\{n\}\{1-x^n\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(大连理工大学2023年数学分析考研试题) [函数极限 ]
[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 1\}\frac\{m(1-x^n)-n(1-x^m)\}\{(1-x^m)(1-x^n)\}\\\\ =&\lim\_\{x\to 1\}\frac\{m-n-mx^n+nx^m\}\{(1-x)^2(1+x+\cdots+x^\{m-1\})(1+x+\cdots+x^\{n-1\})\}\\\\ =&\frac\{1\}\{mn\}\lim\_\{x\to 1\}\frac\{m-n-mx^n+nx^m\}\{(1-x)^2\}\\\\ \xlongequal\{\tiny\mbox\{L'Hospital\}\}& \frac\{1\}\{mn\}\lim\_\{x\to 1\}\frac\{-mnx^\{n-1\}+nmx^\{m-1\}\}\{2(x-1)\} =\frac\{1\}\{2\}\lim\_\{x\to 1\}\frac\{x^\{m-1\}-x^\{n-1\}\}\{x-1\}\\\\ \xlongequal\{\tiny\mbox\{L'Hospital\}\}& \frac\{1\}\{2\}\lim\_\{x\to 1\}[(m-1)x^\{m-2\}-(n-1)x^\{n-2\}] =\frac\{m-n\}\{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](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
100、 (2)、 $\displaystyle \lim\_\{x\to 0\}(\cos x)^\{x^\{-2\}\}=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (电子科技大学2023年数学分析考研试题) [函数极限 ]
[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\[\lim\_\{x\to 0\}\frac\{1\}\{x^2\}\ln \cos x\right\] \xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\} \mathrm\{exp\}\left\[\lim\_\{x\to 0\}\frac\{1\}\{x^2\}(\cos x-1)\right\] \xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\} \mathrm\{e\}^\{-\frac\{1\}\{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](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
101、 (3)、 $\displaystyle \lim\_\{x\to 0\}\frac\{x^2\}\{\int\_\{\cos x\}^1\mathrm\{e\}^\{-\tau^2\}\mathrm\{ d\} \tau\}=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (电子科技大学2023年数学分析考研试题) [函数极限 ]
[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\}\xlongequal\{\tiny\mbox\{L'Hospital\}\} \lim\_\{x\to 0\}\frac\{2x\}\{-\mathrm\{e\}^\{-\cos x^2\}\cdot (-\sin x)\}=2\mathrm\{e\}. \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)/
---
102、 1、 计算极限
\begin\{aligned\} \lim\_\{x\to 0\}\frac\{\sqrt\{1+x\sin x\}-\sqrt\{\cos x\}\}\{x\tan x\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(东北大学2023年数学分析考研试题) [函数极限 ]
[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\}\xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\}&\lim\_\{x\to 0\}\frac\{1+x\sin x-\cos x\}\{x^2\}\cdot \frac\{1\}\{\sqrt\{1+x\sin x\}+\sqrt\{\cos x\}\}\\\\ =&\frac\{1\}\{2\}\lim\_\{x\to 0\}\left(\frac\{\sin x\}\{x\}+\frac\{1-\cos x\}\{x^2\}\right) \xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\} \frac\{1\}\{2\}\left(1+\frac\{1\}\{2\}\right)=\frac\{3\}\{4\}. \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)/
---
103、 (2)、 求 $\displaystyle \lim\_\{x\to+\infty\}\left(\sqrt[6]\{x^6+x^5\}-\sqrt[6]\{x^6-x^5\}\right)$. (东北师范大学2023年数学分析考研试题) [函数极限 ]
[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 +\infty\} x\left(\sqrt[6]\{1+\frac\{1\}\{x\}\}-\sqrt[6]\{1-\frac\{1\}\{x\}\}\right)\\\\ =&\lim\_\{t\to 0\}\frac\{\sqrt[6]\{1+t\}-\sqrt[6]\{1-t\}\}\{t\} =2(\sqrt[6]\{1+t\})'|\_\{t=0\} =\frac\{1\}\{3\}. \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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---
104、 5、 设 $\displaystyle f(x)$ 在 $\displaystyle (a,b)$ 上严格单调, $\displaystyle x\_n\in (a,b)$. 证明: 如果 $\displaystyle \lim\_\{n\to\infty\}f(x\_n)=f(a)$, 则 $\displaystyle \lim\_\{n\to\infty\}x\_n=a$. (东北师范大学2023年数学分析考研试题) [函数极限 ]
[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) / 用反证法. 若 $\displaystyle \lim\_\{n\to\infty\}x\_n=a$ 不成立, 则 $\displaystyle \exists\ \varepsilon\_0 > 0, n\_k,\mathrm\{ s.t.\}$
\begin\{aligned\} &|x\_\{n\_k\}-a|\geq \varepsilon\_0\Rightarrow x\_\{n\_k\} > a+\varepsilon\_0 \stackrel\{f\mbox\{严\}\nearrow \}\{\Rightarrow\}f(x\_\{n\_k\}) > f(a+\varepsilon\_0)\\\\ \stackrel\{k\to\infty\}\{\Rightarrow\}&f(a)\xlongequal\{\tiny\mbox\{题设\}\} \lim\_\{n\to\infty\}f(x\_n) \stackrel\{\mbox\{子列\}\}\{=\}\lim\_\{k\to\infty\}f(x\_\{n\_k\})\geq f(a+\varepsilon) > f(a). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这是一个矛盾. 故有结论.跟锦数学微信公众号. [在线资料](
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---
105、 2、 求 $\displaystyle \lim\_\{x\to 0\}\left\[\cos x+\ln(1+x^2)\right\]^\frac\{1\}\{\sin^2x\}$. (东南大学2023年数学分析考研试题) [函数极限 ]
[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\to 0\}\frac\{\ln [\cos x+\ln (1+x^2)]\}\{\sin^2 x\}\right\\}\\\\ \xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\}& \mathrm\{exp\}\left\\{\lim\_\{x\to 0\}\frac\{\cos x+\ln (1+x^2)-1\}\{x^2\}\right\\} =\mathrm\{exp\}\left\[-\frac\{1\}\{2\}+1\right\]=\sqrt\{\mathrm\{e\}\}. \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)/
---
106、 2、 取定常数 $\displaystyle a,b$, 使得极限
\begin\{aligned\} \lim\_\{x\to 0\}\frac\{ax\cos x-b\sin x\}\{x^3\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
存在, 并求极限. (复旦大学2023年分析(第6,7,8,9,10题没做)考研试题) [函数极限 ]
[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\} ax\cos x\xlongequal[\tiny\mbox\{展开\}]\{\tiny\mbox\{Taylor\}\} ax\left\[1-\frac\{x^2\}\{2\}+o(x^3)\right\], b\sin x=b\left\[x-\frac\{x^3\}\{3!\}+o(x^4)\right\] \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\} \mbox\{原式\}=\lim\_\{x\to 0\}\frac\{(a-b)x+\left(-\frac\{a\}\{2\}+\frac\{b\}\{6\}\right)x^3\}\{x^3\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故当且仅当 $\displaystyle b=a$ 时, 原式存在, 且等于 $\displaystyle -\frac\{a\}\{3\}$.跟锦数学微信公众号. [在线资料](
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---
107、 1、 计算题 (每题 10 分, 共 90 分). (1)、 计算 $\displaystyle \lim\_\{x\to+\infty\}\sin\frac\{3\}\{x\}\ln(1+2^x)$. (广西大学2023年数学分析考研试题) [函数极限 ]
[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\}&\xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\} \lim\_\{x\to+\infty\}\frac\{3\}\{x\}\ln(1+2^x) \xlongequal\{\tiny\mbox\{L'Hospital\}\} 3\lim\_\{x\to+\infty\}\frac\{2^x\ln 2\}\{1+2^x\}=3\ln 2. \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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---
108、 2、 叙述题. 每题 5 分, 共 20 分. (1)、 叙述’$x\to x\_0^+, f(x)\to A$‘的否定命题. (河海大学2023年数学分析考研试题) [函数极限 ]
[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) / $\displaystyle \exists\ \varepsilon\_0 > 0,\ \forall\ \delta > 0, \exists\ x\_0 < x < x\_0+\delta,\mathrm\{ s.t.\} |f(x)-A|\geq \varepsilon\_0$.跟锦数学微信公众号. [在线资料](
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---
109、 3、 计算题. 每题 10 分, 共 50 分. (1)、 计算极限 $\displaystyle \lim\_\{x\to 0\}\frac\{\mathrm\{e\}^x\sin x-x(1+x)\}\{(1+3x)^\frac\{1\}\{x\}(1-\cos x)\arctan x\}$. (河海大学2023年数学分析考研试题) [函数极限 ]
[纸质资料](
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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) /
\begin\{aligned\} \mbox\{原式\}&\xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\} \lim\_\{x\to 0\}\frac\{\mathrm\{e\}^x\sin x-x(1+x)\}\{\left\[(1+3x)^\frac\{1\}\{3x\}\right\]^3\cdot\frac\{x^2\}\{2\}\cdot x\} =\frac\{2\}\{\mathrm\{e\}^3\}\lim\_\{x\to 0\}\frac\{\mathrm\{e\}^x\sin x-x(1+x)\}\{x^3\}\\\\ &\xlongequal[\tiny\mbox\{展开\}]\{\tiny\mbox\{Taylor\}\} \frac\{2\}\{\mathrm\{e\}^3\}\lim\_\{x\to 0\}\frac\{\left\[1+x+\frac\{x^2\}\{2\}+o(x^2)\right\]\left\[x-\frac\{x^3\}\{6\}+o(x^6)\right\]-x(1+x)\}\{x^3\}\\\\ &=\frac\{2\}\{\mathrm\{e\}^3\}\left(-\frac\{1\}\{6\}+\frac\{1\}\{2\}\right)=\frac\{2\}\{3\mathrm\{e\}^3\}. \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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---
110、 1、 计算题 (每小题 12 分, 共 72 分). (1)、 已知 $\displaystyle f(x)$ 二阶可导, $\displaystyle \lim\_\{x\to 0\}\left\[1+x+\frac\{f(x)\}\{x\}\right\]^\frac\{1\}\{x\}=\mathrm\{e\}^3$, 求 $\displaystyle \lim\_\{x\to 0\}\left\[1+\frac\{f(x)\}\{x\}\right\]^\frac\{1\}\{x\}$. (河南大学2023年数学分析考研试题) [函数极限 ]
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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) /
\begin\{aligned\} &\mathrm\{e\}^3=\lim\_\{x\to 0\}\left\[1+x+\frac\{f(x)\}\{x\}\right\]^\frac\{1\}\{x\} =\mathrm\{exp\}\left\\{\lim\_\{x\to 0\}\frac\{\ln \left\[1+x+\frac\{f(x)\}\{x\}\right\]\}\{x\}\right\\}\\\\ \Rightarrow& 3=\lim\_\{x\to 0\}\frac\{\ln \left\[1+x+\frac\{f(x)\}\{x\}\right\]\}\{x\}\\\\ \Rightarrow&\ln \left\\{\lim\_\{x\to 0\}\left\[1+x+\frac\{f(x)\}\{x\}\right\]\right\\} =\lim\_\{x\to 0\}\frac\{\ln \left\[1+x+\frac\{f(x)\}\{x\}\right\]\}\{x\}\cdot x=3\cdot 0=0\\\\ \Rightarrow&\lim\_\{x\to 0\}\left\[1+x+\frac\{f(x)\}\{x\}\right\]=1 \Rightarrow\lim\_\{x\to 0\}\frac\{f(x)\}\{x\}=0\\\\ \Rightarrow&\left\\{\begin\{array\}\{llllllllllll\} f(0)=\lim\_\{x\to 0\}f(x) =\lim\_\{x\to 0\}\frac\{f(x)\}\{x\}\cdot x=0\cdot 0=0\\\\ f'(0)=\lim\_\{x\to 0\}\frac\{f(x)-f(0)\}\{x\} =\lim\_\{x\to 0\}\frac\{f(x)\}\{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\}
进一步, 由上式第 2 行知
\begin\{aligned\} 3&=\lim\_\{x\to 0\}\frac\{\ln \left\[1+x+\frac\{f(x)\}\{x\}\right\]\}\{x\} =\lim\_\{x\to 0\}\frac\{x+\frac\{f(x)\}\{x\}\}\{x\}\left(\ln (1+s)\sim s, s\to 0\right)\\\\ &=1+\lim\_\{x\to 0\}\frac\{f(x)\}\{x^2\} =1+\lim\_\{x\to 0\}\frac\{f'(x)\}\{2x\}\left(\mbox\{L'Hospital 法则\}\right)\\\\ &=1+\frac\{1\}\{2\}\lim\_\{x\to 0\}\frac\{f'(x)-f'(0)\}\{x\} =1+\frac\{1\}\{2\}f''(0). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle f''(0)=4$. 最终,
\begin\{aligned\} &\lim\_\{x\to 0\}\frac\{f(x)\}\{x^2\} =\lim\_\{x\to 0\}\frac\{f'(x)\}\{2x\}=\frac\{f''(0)\}\{2\}=2,\\\\ \mbox\{原式\}=&\mathrm\{exp\}\left\\{\lim\_\{x\to 0\}\frac\{1\}\{x\}\ln\left\[1+\frac\{f(x)\}\{x\}\right\]\right\\} \xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\} \mathrm\{exp\}\left\\{\lim\_\{x\to 0\}\frac\{1\}\{x\}\cdot \frac\{f(x)\}\{x\}\right\\}=\mathrm\{e\}^2. \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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---
111、 1、 求 $\displaystyle \lim\_\{x\to 0\}\left(\frac\{1\}\{x\}-\frac\{1\}\{\mathrm\{e\}^x-1\}\right)$. (黑龙江大学2023年数学分析考研试题) [函数极限 ]
[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\}\frac\{\mathrm\{e\}^x-1-x\}\{x(\mathrm\{e\}^x-1)\} \xlongequal[\tiny\mbox\{展开\}]\{\tiny\mbox\{Taylor\}\} \lim\_\{x\to 0\}\frac\{\frac\{x^2\}\{2\}+o(x^2)\}\{x\cdot x\}=\frac\{1\}\{2\}. \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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---
112、 1、 求极限. (1)、 求 $\displaystyle \lim\_\{x\to 0\}\frac\{\mathrm\{e\}^\{\alpha x^2\}\cos^\{2\alpha\}x-1\}\{x^2\}$. (华东理工大学2023年数学分析考研试题) [函数极限 ]
[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) / 若 $\displaystyle \alpha=0$, 则极限 $\displaystyle =0$. 若 $\displaystyle \alpha\neq 0$, 则
\begin\{aligned\} \mbox\{原式\}=&\lim\_\{x\to 0\}\left\[\frac\{\mathrm\{e\}^\{\alpha x^2\}(\cos^\{2\alpha\}x-1)\}\{x^2\}+\frac\{\mathrm\{e\}^\{\alpha x^2\}-1\}\{x^2\}\right\] \xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\} \lim\_\{x\to 0\}\frac\{\cos^\{2\alpha\}x-1\}\{x^2\}+\alpha\\\\ \xlongequal[\tiny\mbox\{展开\}]\{\tiny\mbox\{Taylor\}\}&\lim\_\{x\to 0\}\frac\{\left\[1-\frac\{x^2\}\{2\}+o(x^2)\right\]^\{2\alpha\}-1\}\{x^2\}+\alpha\\\\ \xlongequal[\tiny\mbox\{展开\}]\{\tiny\mbox\{Taylor\}\}& \lim\_\{x\to 0\}\frac\{2\alpha\left(-\frac\{x^2\}\{2\}\right)+o(x^2)\}\{x^2\}+\alpha=0. \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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---
113、 7、 设 $\displaystyle f(x)$ 在 $\displaystyle (0,+\infty)$ 上可导, 且 $\displaystyle \lim\_\{x\to+\infty\}[f(x)+xf'(x)\ln x]=a$. 证明: $\displaystyle \lim\_\{x\to+\infty\}f(x)=a$. (华东理工大学2023年数学分析考研试题) [函数极限 ]
[纸质资料](
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https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \lim\_\{x\to+\infty\}f(x)=&\lim\_\{x\to+\infty\}\frac\{f(x)\ln x\}\{\ln x\} \xlongequal\{\tiny\mbox\{L'Hospital\}\} \lim\_\{x\to+\infty\}\frac\{f'(x)\ln x+f(x)\frac\{1\}\{x\}\}\{\frac\{1\}\{x\}\}\\\\ =&\lim\_\{x\to+\infty\}\left\[f(x)+xf'(x)\ln x\right\]=a. \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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---
114、 1、 计算极限
\begin\{aligned\} \lim\_\{x\to 0^+\}\frac\{\sqrt\{1-\mathrm\{e\}^\{-x\}\}-\sqrt\{1-\cos x\}\}\{\sqrt\{\sin x\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(华东师范大学2023年数学分析考研试题) [函数极限 ]
[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\}\frac\{\cos x-\mathrm\{e\}^\{-x\}\}\{\sqrt\{\sin x\}\}\cdot \frac\{1\}\{\sqrt\{1-\mathrm\{e\}^\{-x\}+\sqrt\{1-\cos x\}\}\}\\\\ \xlongequal[\tiny\mbox\{展开\}]\{\tiny\mbox\{Taylor\}\}&\lim\_\{x\to 0\}\frac\{[1+o(x)]-[1-x+o(x)]\}\{\sqrt\{x\}\}\cdot \frac\{1\}\{\sqrt\{1-\mathrm\{e\}^\{-x\}\}+\sqrt\{1-\cos x\}\}\\\\ =&\lim\_\{x\to 0\}\frac\{\sqrt\{x\}\}\{\sqrt\{1-\mathrm\{e\}^\{-x\}\}+\sqrt\{1-\cos x\}\} =\lim\_\{x\to 0\}\frac\{1\}\{\sqrt\{\frac\{1-\mathrm\{e\}^\{-x\}\}\{x\}\}+\sqrt\{\frac\{2\sin^2\frac\{x\}\{2\}\}\{x\}\}\}\\\\ =&\frac\{1\}\{1+0\}=1. \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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---
115、 1、 求极限 $\displaystyle \lim\_\{x\to+\infty\}\left(\sqrt[m]\{x^m+x^\{m-1\}\}-\sqrt[m]\{x^m-x^\{m-1\}\}\right)$. (华南理工大学2023年数学分析考研试题) [函数极限 ]
[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+\infty\}x\left\[\sqrt[m]\{1+\frac\{1\}\{x\}\}-\sqrt[m]\{1-\frac\{1\}\{x\}\}\right\] =\lim\_\{t\to 0\}\frac\{\sqrt[m]\{1+t\}-\sqrt[m]\{1-t\}\}\{t\}\\\\ =&\lim\_\{t\to 0\} \left\[\frac\{\sqrt[m]\{1+t\}-1\}\{t\}-\frac\{\sqrt[m]\{1-t\}-1\}\{-t\}\right\]\\\\ =&\left.\frac\{\mathrm\{ d\}\}\{\mathrm\{ d\} t\}\right|\_\{t=0\} (1+t)^\frac\{1\}\{m\} +\left.\frac\{\mathrm\{ d\}\}\{\mathrm\{ d\} s\}\right|\_\{s=0\} (1+s)^\frac\{1\}\{m\}=\frac\{2\}\{m\}. \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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发表于 2023-3-5 08:41:07
