## 张祖锦2023年数学专业真题分类70天之第07天
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139、 2、 求极限 $\displaystyle \lim\_\{x\to 0\}\frac\{(1+x)^\frac\{1\}\{x\}-\mathrm\{e\}\}\{\sqrt\{1+x\}-1\}$. (太原理工大学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 0\}\frac\{(1+x)^\frac\{1\}\{x\}-\mathrm\{e\}\}\{x\}(\sqrt\{1+x\}+1) =2\lim\_\{x\to 0\} \frac\{\mathrm\{e\}^\frac\{\ln(1+x)\}\{x\}-\mathrm\{e\}\}\{x\}\\\\ &\xlongequal[\frac\{\ln(1+x)\}\{x\} < \xi\_x < 1]\{\mbox\{Lagrange\}\}\mathrm\{e\}^\{\xi\_x\}\frac\{\frac\{\ln(1+x)\}\{x\}-1\}\{x\} =2\mathrm\{e\}\lim\_\{x\to 0\}\frac\{\ln(1+x)-x\}\{x^2\}\\\\ &\xlongequal\{\tiny\mbox\{L'Hospital\}\} 2\lim\_\{x\to 0\}\frac\{\frac\{1\}\{1+x\}-1\}\{2x\} =-\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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140、 1、 计算题. (1)、 $\displaystyle \lim\_\{x\to-\infty\}\left(\frac\{a\_1^x+\cdots+a\_n^x\}\{n\}\right)^\frac\{1\}\{x\}$, 其中 $\displaystyle a\_1,\cdots,a\_n$ 均为正数; (天津大学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) / 设 $\displaystyle \min\_ia\_i=\left\\{a\_\{i\_1\},\cdots, a\_\{i\_s\}\right\\}$, $\displaystyle S=\left\\{i\_1,\cdots,i\_s\right\\}$, 则
\begin\{aligned\} \mbox\{原式\}=&\mathrm\{exp\}\left\[\lim\_\{x\to-\infty\}\frac\{\ln\frac\{a\_1^x+\cdots+a\_n^x\}\{n\}\}\{n\}\right\]\\\\ \xlongequal\{\tiny\mbox\{L'Hospital\}\}& \mathrm\{exp\}\left\[\lim\_\{x\to-\infty\}\frac\{1\}\{\frac\{a\_1^x+\cdots+a\_n^x\}\{n\}\}\cdot\frac\{a\_1^x\ln a\_1+\cdots+a\_n^x\ln a\_n\}\{n\}\right\]\\\\ \stackrel\{\div a\_\{i\_1\}^x\}\{=\}&\mathrm\{exp\}\left\[\lim\_\{x\to-\infty\}\frac\{\displaystyle s\ln a\_\{i\_1\}+\sum\_\{i\not\in S\} \left(\frac\{a\_i\}\{a\_\{i\_1\}\}\right)^\{x\ln a\_i\}\}\{\displaystyle s+\sum\_\{i\not\in S\}\left(\frac\{a\_i\}\{a\_\{i\_1\}\}\right)^x\}\right\]\\\\ =&\mathrm\{exp\}\left\[\frac\{s\ln a\_\{i\_1\}\}\{s\}\right\]=a\_\{i\_1\}=\min\_\{1\leq i\leq n\}a\_i\left(i\not\in S\Rightarrow \frac\{a\_i\}\{a\_\{i\_1\}\} > 1\right). \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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141、 8、 设 $\displaystyle \beta > 0$. 讨论 $\displaystyle \alpha$ 的取值, 使得 $\displaystyle \lim\_\{x\to 0^+\}\frac\{1\}\{x^\alpha\}\int\_0^x\sin\frac\{1\}\{\beta t\}\mathrm\{ d\} t$ 存在, 并求其值. (天津大学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) / 当且仅当 $\displaystyle \alpha < 2$ 时, 极限存在, 且等于 $\displaystyle 0$. 由
\begin\{aligned\} &\int\_0^x \sin\frac\{1\}\{\beta t\}\mathrm\{ d\} t\stackrel\{\beta t=s\}\{=\} \frac\{1\}\{\beta\}\int\_0^\{\beta x\}\sin \frac\{1\}\{s\}\mathrm\{ d\} s =\frac\{1\}\{\beta\}\int\_0^\{\beta x\}s^2\cdot \frac\{1\}\{s^2\}\cos\frac\{1\}\{s\}\mathrm\{ d\} s\\\\ =&\frac\{1\}\{\beta\}\int\_0^\{\beta x\} s^2\mathrm\{ d\} \cos\frac\{1\}\{s\} \xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\} \frac\{1\}\{\beta\}\left\[\beta^2x^2 \cos\frac\{1\}\{\beta x\}-\int\_0^\{\beta x\}2s\cos\frac\{1\}\{s\}\mathrm\{ d\} s\right\]\qquad(I) \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\} \left|\int\_0^x \sin\frac\{1\}\{\beta t\}\mathrm\{ d\} t\right|\leq \beta x^2+\int\_0^\{\beta x\}2s\mathrm\{ d\} s=(\beta+\beta^2)x^2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故当 $\displaystyle \alpha < 2$ 时, 由迫敛性知原式 $\displaystyle =0$. 但当 $\displaystyle \alpha\geq 2$ 时, 极限不存在. 事实上, 由 $\displaystyle \frac\{1\}\{x^\alpha\}=\frac\{1\}\{x^\{\alpha-2\}\}\cdot \frac\{1\}\{x^2\}$ 知仅需验证 $\displaystyle \lim\_\{x\to 0^+\}\frac\{1\}\{x^2\}\int\_0^x \sin\frac\{1\}\{\beta t\}\mathrm\{ d\} t$ 不存在. 按 $\displaystyle (I)$ 写出
\begin\{aligned\} \frac\{1\}\{x^2\}\int\_0^x \sin\frac\{1\}\{\beta t\}\mathrm\{ d\} t =\beta\cos\frac\{1\}\{\beta x\}-\frac\{1\}\{\beta x^2\}\int\_0^\{\beta x\}2s\cos \frac\{1\}\{s\}\mathrm\{ d\} s. \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\} &\int\_0^\{\beta x\}2s\cos \frac\{1\}\{s\}\mathrm\{ d\} s =\int\_0^\{\beta x\}(-2s^3)\left(-\frac\{1\}\{s^2\}\right)\cos\frac\{1\}\{s\}\mathrm\{ d\} s\\\\ =&\int\_0^\{\beta x\}(-2s^3)\mathrm\{ d\} \sin\frac\{1\}\{s\} =-2\left\[\beta^3x^3\sin\frac\{1\}\{\beta x\}-\int\_0^\{\beta x\}3s^2\sin \frac\{1\}\{s\}\mathrm\{ d\} s\right\] \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\} \left|\frac\{1\}\{x^2\}\int\_0^\{\beta x\}2s\cos \frac\{1\}\{s\}\mathrm\{ d\} s\right|\leq 2\beta^3 x^2 +\frac\{1\}\{x^2\}\int\_0^\{\beta x\}3s^2\mathrm\{ d\} s\xrightarrow\{x\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\}
故仅需验证 $\displaystyle \lim\_\{x\to 0^+\}\cos\frac\{1\}\{\beta x\}$ 不存在. 而这可由
\begin\{aligned\} \cos\frac\{1\}\{\beta\cdot\frac\{1\}\{2n\pi\beta\}\}=1, \cos\frac\{1\}\{\beta\cdot\frac\{1\}\{\left(2n\pi+\frac\{\pi\}\{2\}\right)\beta\}\}=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及归结原理得到. 解毕.跟锦数学微信公众号. [在线资料](
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142、 (2)、 求极限 $\displaystyle \lim\_\{x\to 0\}\frac\{\ln \left(\mathrm\{e\}^\{\tan x\}+\sqrt[3]\{1-\cos x\}\right)+\tan x\}\{\arcsin\left(2023\sqrt[3]\{1-\cos x\}\right)\}$. (武汉大学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\} \sqrt[3]\{1-\cos x\}\sim\sqrt[3]\{\frac\{x^2\}\{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
\begin\{aligned\} \mbox\{原式\}&\xlongequal[\tiny\mbox\{展开\}]\{\tiny\mbox\{Taylor\}\} \lim\_\{x\to 0\}\frac\{\left\[\begin\{array\}\{c\}\mathrm\{e\}^\{\tan x\}-1+\sqrt[3]\{1-\cos x\}\\\\ +\frac\{\left(\mathrm\{e\}^\{\tan x\}-1+\sqrt[3]\{1-\cos x\}\right)^2\}\{2\}+o(x^2)\end\{array\}\right\]-\tan x\}\{2023\sqrt[3]\{1-\cos x\}\}\\\\ &\stackrel\{\mathrm\{e\}^s-1-s\sim\frac\{s^2\}\{2\}, s\to 0\}\{=\}\frac\{1\}\{2023\}. \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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143、 1、 计算题. (1)、 求极限 $\displaystyle \lim\_\{x\to 0\}\frac\{\mathrm\{e\}^x-\sqrt[3]\{1+3x\}\}\{\ln (1+x^2)\}$. (武汉理工大学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) / 由 $\displaystyle (1+t)^\frac\{1\}\{3\}=1+\frac\{t\}\{3\}-\frac\{t^2\}\{9\}+o(t^2), t\to 0$ 知
\begin\{aligned\} \mbox\{原式\}\xlongequal[\tiny\mbox\{展开\}]\{\tiny\mbox\{Taylor\}\}&\lim\_\{x\to 0\}\frac\{\left\[1+x+\frac\{x^2\}\{2\}+o(x^2)\right\]-\left\[1+\frac\{3x\}\{3\}-\frac\{(3x)^2\}\{9\}+o(x^2)\right\]\}\{x^2\}\\\\ =&\frac\{1\}\{2\}+1=\frac\{3\}\{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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144、 (2)、 $\displaystyle \lim\_\{x\to+\infty\}\left(\sqrt[3]\{x^3-x^2\}-\sqrt[3]\{x^3+x\}\right)=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (西安交通大学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[3]\{1-\frac\{1\}\{x\}\}-\sqrt[3]\{1+\frac\{1\}\{x^2\}\}\right) =\lim\_\{t\to 0\}\frac\{\sqrt[3]\{1-t\}-\sqrt[3]\{1+t^2\}\}\{t\}\\\\ \xlongequal[\tiny\mbox\{展开\}]\{\tiny\mbox\{Taylor\}\}&s\lim\_\{t\to 0\}\frac\{\left\[1-\frac\{t\}\{3\}+o(t)\right\]-\left\[1+\frac\{t^2\}\{3\}+o(t^2)\right\]\}\{t\} =-\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\}
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145、 (2)、 设函数 $\displaystyle f(x)$ 在 $\displaystyle a$ 的邻域内有定义, 在 $\displaystyle a$ 处可导且 $\displaystyle f(a) > 0$. 计算
\begin\{aligned\} \lim\_\{x\to a\}\frac\{f(x)^\{f(x)\}-f(a)^\{f(x)\}\}\{x-a\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(西安交通大学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) / 设 $\displaystyle g(t)=t^\{f(x)\}$, 则
\begin\{aligned\} \mbox\{原式\}=&\lim\_\{x\to a\}\frac\{g\left(f(x)\right)-g\left(f(a)\right)\}\{x-a\} \xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\} \lim\_\{x\to a\}g'(\xi\_x)\frac\{f(x)-f(a)\}\{x-a\}\\\\ =&\lim\_\{x\to a\}f(x) \xi\_x^\{f(x)-1\} f'(a) =f(a)^\{f(a)\}f'(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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146、 1、 (20 分) 求下列极限. (1)、 $\displaystyle \lim\_\{x\to 0\}\frac\{(1+\sin^2x)^\{1902\}-(\cos x)^\{2022\}\}\{\tan^2x\}$. (西北大学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\{展开\}]\{\tiny\mbox\{Taylor\}\}& \lim\_\{x\to 0\}\frac\{\left\[1+1902\sin^2x+o(x^2)\right\]-\left\[1+2022(\cos x-1)+o(x^2)\right\]\}\{x^2\}\\\\ =&1902-2022\left(-\frac\{1\}\{2\}\right)=2913. \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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147、 1、 (20 分) 计算极限
\begin\{aligned\} \lim\_\{x\to 0\}\frac\{\int\_0^x t\cos t\mathrm\{ d\} t-1+\cos x\}\{\sqrt\{1+x\tan x\}-\sqrt\{1+x\sin x\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(西南财经大学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\{\int\_0^x t\cos t\mathrm\{ d\} t-1+\cos x\}\{x(\tan x-\sin x)\} \left(\sqrt\{1+x\tan x\}+\sqrt\{1+x\sin x\}\right)\\\\ =&2\lim\_\{x\to 0\}\frac\{\int\_0^x t\cos t\mathrm\{ d\} t-1+\cos x\}\{x\tan x(1-\cos x)\} \xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\} 4\lim\_\{x\to 0\}\frac\{\int\_0^x t\cos t\mathrm\{ d\} t-1+\cos x\}\{x^4\}\\\\ \xlongequal\{\tiny\mbox\{L'Hospital\}\}&4\lim\_\{x\to 0\}\frac\{x\cos x-\sin x\}\{4x^3\} \xlongequal\{\tiny\mbox\{L'Hospital\}\} 4\lim\_\{x\to 0\}\frac\{-x\sin x\}\{12x^2\}=-\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\}
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148、 (2)、 (10 分) 求极限 $\displaystyle \lim\_\{x\to 0\}\frac\{\sin \sin x+\sin 2x\}\{\tan x-3\arctan 2x\}$. (西南大学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\}\xlongequal\{\tiny\mbox\{L'Hospital\}\} \lim\_\{x\to 0\}\frac\{\cos\sin x\cdot \cos x+2\cos 2x\}\{\sec^2x-3\frac\{1\}\{1+4x^2\}\cdot 2\} =\frac\{1+2\}\{1-6\}=-\frac\{3\}\{5\}. \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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149、 2、 (10 分) 求 $\displaystyle \lim\_\{x\to 0\}\frac\{1-\cos x\cos 2x\cos 3x\}\{1-\cos x\}$. (西南交通大学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-\cos x+\cos x(1-\cos 2x\cos 3x)\}\{\frac\{x^2\}\{2\}\}\\\\ =&1+\lim\_\{x\to 0\}\frac\{1-\cos 2x+\cos 2x(1-\cos 3x)\}\{\frac\{x^2\}\{2\}\}\\\\ \xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\}&1=\frac\{\frac\{2^2\}\{2\}\}\{\frac\{1\}\{2\}\}+\frac\{\frac\{3^2\}\{2\}\}\{\frac\{1\}\{2\}\}=1+4+9=14. \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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https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
150、 1、 求极限. (1)、 $\displaystyle \lim\_\{x\to 0\}\frac\{\mathrm\{e\}^x\sin x-x(1+x)\}\{x^3\}$; (湘潭大学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\{\left\[1+x+\frac\{x^2\}\{2\}+o(x^2)\right\]\left\[x-\frac\{x^3\}\{6\}+o(x^3)\right\]-x-x^2\}\{x^3\}\\\\ &=-\frac\{1\}\{6\}+\frac\{1\}\{2\}=\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\}
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---
151、 (2)、 $\displaystyle \lim\_\{x\to 0\}\frac\{\int\_0^x \sin t\mathrm\{ d\} t\}\{x\sin x\}$; (湘潭大学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\{\int\_0^x \sin t\mathrm\{ d\} t\}\{x^2\} \xlongequal\{\tiny\mbox\{L'Hospital\}\} \lim\_\{x\to 0\}\frac\{\sin x\}\{2x\}=\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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---
152、 7、 若 $\displaystyle f(x)$ 在 $\displaystyle (a,+\infty)$ 内可导 (其中 $\displaystyle a$ 是给定的实数), 且 $\displaystyle \lim\_\{x\to+\infty\}f'(x)=A$. 试证明: $\displaystyle \lim\_\{x\to+\infty\}\frac\{f(x)\}\{x\}=A$. (湘潭大学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 $\displaystyle \lim\_\{x\to+\infty\}f'(x)=0$ 知
\begin\{aligned\} \forall\ \varepsilon > 0,\exists\ X > 0,\mathrm\{ s.t.\} \forall\ x\geq X, |f'(x)| < \frac\{\varepsilon\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
又由 $\displaystyle \lim\_\{x\to+\infty\}\frac\{f(X)\}\{x\}=0$ 知
\begin\{aligned\} \exists\ X' > X,\mathrm\{ s.t.\} \forall\ x\geq X', \left|\frac\{f(X)\}\{x\}\right| < \frac\{\varepsilon\}\{2\}. \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 $\displaystyle x\geq X'$ 时,
\begin\{aligned\} \left|\frac\{f(x)\}\{x\}\right|\leq&\left|\frac\{f(X)\}\{x\}\right|+\frac\{\left|f(x)-f(X)\right|\}\{x\}\\\\ \xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\}&\left|\frac\{f(X)\}\{x\}\right|+\frac\{|f'(\xi\_x)|(x-X)\}\{x\} < \frac\{\varepsilon\}\{2\}+\frac\{\varepsilon\}\{2\}=\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 $\displaystyle \lim\_\{x\to+\infty\}\frac\{f(x)\}\{x\}=0$. 一言以蔽之, 这就是函数极限的 L'Hospital 法则.跟锦数学微信公众号. [在线资料](
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---
153、 2、 (15 分) 计算极限. (1)、 (8 分) $\displaystyle \lim\_\{x\to 0\}\frac\{\mathrm\{e\}^x-(1+2x)^\frac\{1\}\{2\}\}\{\ln(1+x^2)\}$; (新疆大学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\{\mathrm\{e\}^x-\mathrm\{e\}^\{\frac\{1\}\{2\}\ln (1+2x)\}\}\{x^2\} \xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\} \lim\_\{x\to 0\}\frac\{\mathrm\{e\}^\{\xi\_x\}\left\[x-\frac\{\ln(1+2x)\}\{2\}\right\]\}\{x^2\}\\\\ &=\frac\{1\}\{2\}\lim\_\{x\to 0\}\frac\{2x-\ln (1+x)\}\{x^2\} \xlongequal[\tiny\mbox\{展开\}]\{\tiny\mbox\{Taylor\}\} \frac\{1\}\{2\}\lim\_\{x\to 0\}\frac\{\frac\{(2x)^2\}\{2\}+o(x^2)\}\{x^2\} =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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---
154、 (2)、 (7 分) $\displaystyle \lim\_\{x\to 0^+\}(\sin x)^\frac\{1\}\{1+\ln x\}$. (新疆大学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\[\lim\_\{x\to 0^+\}\frac\{1\}\{1+\ln x\}\ln \sin x\right\] \xlongequal\{\tiny\mbox\{L'Hospital\}\} \mathrm\{exp\}\left\[\lim\_\{x\to 0^+\}\frac\{\frac\{1\}\{\sin x\}\cos x\}\{\frac\{1\}\{x\}\}\right\]=\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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---
155、 (2)、 求 $\displaystyle \lim\_\{x\to 0\}(1+\sin x)^\frac\{1\}\{x\}$. (云南大学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\}=&\mathrm\{exp\}\left\[\lim\_\{x\to 0\}\frac\{1\}\{x\}\ln (1+\sin x)\right\] \xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\} \mathrm\{exp\}\left\[\lim\_\{x\to 0\}\frac\{\sin x\}\{x\}\right\]=\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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---
156、 2、 填空题 (每题 5 分, 共 30 分). (1)、 极限 $\displaystyle \lim\_\{x\to 1\}\frac\{(1-x^\frac\{1\}\{2\})(1-x^\frac\{1\}\{3\})\cdots(1-x^\frac\{1\}\{n\})\}\{(1-x)^\{n-1\}\}=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (长安大学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\{1\}\{(1-x)^\{n-1\}\}\prod\_\{i=2\}^n (1-x^\frac\{1\}\{i\})\\\\ \xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\}& \lim\_\{x\to 1\}\frac\{1\}\{(1-x)^\{n-1\}\}\prod\_\{i=2\}^n \left\[\frac\{1\}\{i\}\xi\_i^\{\frac\{1\}\{i\}-1\} (1-x)\right\] =\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\}
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---
157、 3、 计算题. (1)、 (10 分) 计算极限 $\displaystyle \lim\_\{x\to 0\}\frac\{1-(\cos x)^\{\sin x\}\}\{x^3\}$. (长安大学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\{1-\mathrm\{e\}^\{\sin x\ln \cos x\}\}\{x^3\} \xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\} \lim\_\{x\to 0\}\frac\{-\sin x\ln \cos x\}\{x^3\}\\\\ \xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\}& \lim\_\{x\to 0\}\frac\{-\sin x(\cos x-1)\}\{x^3\} \xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\} \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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---
158、 1、 (10 分) 求函数极限 $\displaystyle \lim\_\{x\to+\infty\}\left(\sqrt[6]\{x^6+x^5\}-\sqrt[4]\{x^4-x^3\}\right)$. (郑州大学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\[\sqrt[6]\{1+\frac\{1\}\{x\}\}-\sqrt[4]\{1-\frac\{1\}\{x\}\}\right\] =\lim\_\{t\to 0\}\frac\{(1+t)^\frac\{1\}\{6\}-(1+t)^\frac\{1\}\{4\}\}\{t\}\\\\ \xlongequal[\tiny\mbox\{展开\}]\{\tiny\mbox\{Taylor\}\}&\lim\_\{t\to 0\}\frac\{\left\[1+\frac\{t\}\{6\}+o(t^2)\right\]-\left\[1-\frac\{t\}\{4\}+o(t^2)\right\]\}\{t\}=\frac\{5\}\{12\}. \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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---
159、 (2)、 (7 分) 求 $\displaystyle \lim\_\{x\to 1\}\frac\{x^x-x\}\{\ln x-x+1\}$. (中国科学技术大学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\{\mathrm\{e\}^\{x\ln x\}-\mathrm\{e\}^\{\ln x\}\}\{\ln x-x+1\} \xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\} \lim\_\{x\to 1\}\frac\{\mathrm\{e\}^\{\xi\_x\}(x\ln x-\ln x)\}\{\ln x-x+1\}\\\\ &=\lim\_\{x\to 1\}\frac\{\ln x(x-1)\}\{\ln x-(x-1)\} =\lim\_\{t\to 0\}\frac\{t\ln(1+t)\}\{\ln(1+t)-t\}\\\\ &\xlongequal[\tiny\mbox\{展开\}]\{\tiny\mbox\{Taylor\}\} \lim\_\{t\to 0\}\frac\{t[t+o(t)]\}\{\left\[t-\frac\{t^2\}\{2\}+o(t^2)\right\]-t\}=-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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160、 2、 (10 分) 求函数极限 $\displaystyle \lim\_\{x\to 0\}\frac\{(\sin x-\sin\sin x)\sin x\}\{x^4\}$. (中国矿业大学(北京)2023年数学分析考研试题) [函数极限 ]
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https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
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https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{原式\}\xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\} \lim\_\{x\to 0\}\frac\{\cos\xi\_x(x-\sin x)\}\{x^3\}\cdot \frac\{\sin x\}\{x\} \xlongequal[\tiny\mbox\{展开\}]\{\tiny\mbox\{Taylor\}\} \frac\{1\}\{6\}. \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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---
161、 1、 计算题 (每题 10 分, 共 40 分). (1)、 设 $\displaystyle m,n$ 为正整数, 求极限
\begin\{aligned\} \lim\_\{x\to 0\}\frac\{\sqrt[m]\{\cos x\}-\sqrt[n]\{\cos x\}\}\{\sin^2x\}\left(m,n\in\mathbb\{Z\}\_+\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(中南大学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\}\xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\}&\lim\_\{x\to 0\}\frac\{\cos^\frac\{1\}\{m\}x-\cos^\frac\{1\}\{n\}x\}\{x^2\} \xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\} \lim\_\{x\to 0\}\frac\{\cos^\{\xi\_x\}x \ln \cos x\left(\frac\{1\}\{m\}-\frac\{1\}\{n\}\right)\}\{x^2\}\\\\ \xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\}&\left(\frac\{1\}\{m\}-\frac\{1\}\{n\}\right)\lim\_\{x\to 0\}\frac\{\cos x-1\}\{x^2\} \xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\} -\frac\{1\}\{2\}\left(\frac\{1\}\{m\}-\frac\{1\}\{n\}\right) =\frac\{m-n\}\{2mn\}. \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 08:42:21
