## 张祖锦2023年数学专业真题分类70天之第06天
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116、 1、 (40 分) 求极限. (1)、 $\displaystyle \lim\_\{x\to 0\}\frac\{(\tan x)^2(1-\cos x)^2\}\{x(\arcsin x)^3[\ln(1+x)]^2\}$. (华南师范大学2023年数学分析考研试题) [函数极限 ]
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https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
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\begin\{aligned\} \mbox\{原式\}\xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\} \lim\_\{x\to 0\}\frac\{x^2\left(\frac\{x^2\}\{2\}\right)^2\}\{x\cdot x^3\cdot x^2\}=\frac\{1\}\{4\}. \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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117、 3、 (15 分) 设函数 $\displaystyle f(x)$ 在 $\displaystyle \mathring\{U\}(x\_0)$ 内有定义且 $\displaystyle \lim\_\{x\to x\_0\} f(x)=A$ 存在. 证明: (1)、 存在 $\displaystyle \delta\_0 > 0$, 使得 $\displaystyle \forall\ x\in\mathring\{U\}(x\_0;\delta\_0)$, $\displaystyle |f(x)| > \frac\{|A|\}\{2\} > 0$ (函数极限的局部保号性); (2)、 利用函数极限定义及局部保号性, 证明:
\begin\{aligned\} \lim\_\{x\to x\_0\} \tan x=\tan x\_0\left(x\_0\neq k\pi+\frac\{\pi\}\{2\}, k\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\}
(华南师范大学2023年数学分析考研试题) [函数极限 ]
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https://mp.weixin.qq.com/s/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)、 由函数极限的定义知 $\displaystyle \lim\_\{x\to x\_0\} |f(x)|=|A|$, 而
\begin\{aligned\} &\exists\ \delta\_0 > 0,\mathrm\{ s.t.\} \forall\ x\in \mathring\{U\}(x\_0;\delta\_0),\\\\ &\left||f(x)|-|A|\right| < \frac\{|A|\}\{2\} \Rightarrow |f(x)| > \frac\{|A|\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 由 $\displaystyle \tan x$ 的周期性及是奇函数知不妨设 $\displaystyle 0 < x\_0 < \frac\{\pi\}\{2\}$. 此时,
\begin\{aligned\} &\forall\ \varepsilon > 0,\exists\ \delta=\min\left\\{x\_0,\frac\{\frac\{\pi\}\{2\}-x\_0\}\{2\},\varepsilon \cos^2\left(\frac\{\pi\}\{4\}+\frac\{x\_0\}\{2\}\right)\right\\} > 0,\\\\ &\mathrm\{ s.t.\} \forall\ x: x\in \mathring\{U\}(x\_0;\delta),\\\\ &|\tan x-\tan x\_0|\xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\} |\sec^2\xi|\cdot |x-x\_0| =\frac\{1\}\{\cos^2\xi\}|x-x\_0|\\\\ &\leq \frac\{1\}\{\cos^2\left(\frac\{\pi\}\{4\}+\frac\{x\_0\}\{2\}\right)\}|x-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\}
故确有 $\displaystyle \lim\_\{x\to x\_0\} \tan x=\tan x\_0$.跟锦数学微信公众号. [在线资料](
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118、 6、 (10 分) 设函数 $\displaystyle f(x)$ 是定义在 $\displaystyle [a,+\infty)$ 上的单调递减函数. 证明: $\displaystyle \lim\_\{x\to+\infty\}f(x)$ 存在且有限的充要条件是 $\displaystyle f(x)$ 在 $\displaystyle [a,+\infty)$ 上有下界. (华南师范大学2023年数学分析考研试题) [函数极限 ]
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https://mp.weixin.qq.com/s/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)、 $\displaystyle \Rightarrow$: 设 $\displaystyle \lim\_\{x\to+\infty\}f(x)=\ell$ 存在, 则
\begin\{aligned\} \exists\ X > a,\mathrm\{ s.t.\} \forall\ x\geq X, |f(x)-\ell| < 1\Rightarrow f(x) > \ell-1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
又由 $\displaystyle f$ 递减知
\begin\{aligned\} \forall\ x\geq a, f(x)\geq f(\max\left\\{x,X\right\\})\geq \ell-1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle f$ 有下界. (2)、 $\displaystyle \Leftarrow$: 若 $\displaystyle f$ 有下界, 则 $\displaystyle f$ 有下确界 $\displaystyle \ell=\inf\_\{[a,+\infty)\}f$. 按定义,
\begin\{aligned\} \forall\ \varepsilon > 0,\exists\ X\geq a,\mathrm\{ s.t.\} f(X) < \ell+\varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是当 $\displaystyle x\geq X$ 时,
\begin\{aligned\} \ell\leq f(x)\leq f(X) < \ell+\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 $\displaystyle \lim\_\{x\to+\infty\}f(x)=\ell$.跟锦数学微信公众号. [在线资料](
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119、 1、 (15 分) 求极限 $\displaystyle \lim\_\{x\to 0\}\left(-\frac\{\cot x\}\{\mathrm\{e\}^\{-2x\}\}+\frac\{1\}\{\mathrm\{e\}^\{-x\}\sin^2x\}-\frac\{1\}\{x^2\}\right)$. (华中科技大学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\} \mbox\{原式\}=&\lim\_\{x\to 0\}\left(-\frac\{\mathrm\{e\}^x \cos x\}\{\sin x\} +\frac\{\mathrm\{e\}^\{2x\}\}\{\sin^2x\} -\frac\{1\}\{x^2\}\right)\\\\ =&\lim\_\{x\to 0\}\frac\{-x^2\mathrm\{e\}^\{2x\}\sin x\cos x+x^2\mathrm\{e\}^x-\sin^2x\}\{x^2\sin^2x\}\\\\ \xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\}& \lim\_\{x\to 0\}\frac\{-\frac\{x^2\}\{2\}\mathrm\{e\}^\{2x\}\sin 2x+x^2\mathrm\{e\}^x-\frac\{1-\cos 2x\}\{2\}\}\{x^4\}\\\\ \xlongequal[\tiny\mbox\{展开\}]\{\tiny\mbox\{Taylor\}\}&\lim\_\{x\to 0\}\frac\{\boxed\{\begin\{array\}\{c\}-\frac\{x^2\}\{2\}\left\[1+2x+o(x^2)\right\]\left\[2x-\frac\{(2x)^3\}\{3!\}+o(x^3)\right\]\\\\ +x^2\left\[1+x+\frac\{x^2\}\{2\}+o(x^2)\right\] -\frac\{1\}\{2\}\left\[\frac\{(2x)^2\}\{2\}-\frac\{(2x)^4\}\{4!\}+o(x^4)\right\]\end\{array\}\}\}\{x^4\}\\\\ =&-2+\frac\{1\}\{2\}+\frac\{1\}\{3\}=-\frac\{7\}\{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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120、 1、 计算题 (每题 10 分, 共 50 分). (1)、 计算极限 $\displaystyle \lim\_\{x\to+\infty\}\sqrt\{x^3\}\left(\sqrt\{x+1\}+\sqrt\{x-1\}-2\sqrt\{x\}\right)$. (华中师范大学2023年数学分析考研试题) [函数极限 ]
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https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{原式\}&\stackrel\{\frac\{1\}\{x\}=t\}\{=\}\lim\_\{t\to 0\}\frac\{1\}\{t^\frac\{3\}\{2\}\}\left(\sqrt\{\frac\{1\}\{t\}+1\}+\sqrt\{\frac\{1\}\{t\}-1\}-2\sqrt\{\frac\{1\}\{t\}\}\right)\\\\ &=\lim\_\{t\to 0\} \frac\{\sqrt\{1+t\}+\sqrt\{1-t\}-2\}\{t^2\}\\\\ &\xlongequal[\tiny\mbox\{展开\}]\{\tiny\mbox\{Taylor\}\} \lim\_\{t\to 0\}\frac\{\left\[1+\frac\{t\}\{2\}-\frac\{t^2\}\{8\}+o(t^2)\right\]-\left\[1+\frac\{-t\}\{2\}-\frac\{(-t)^2\}\{8\}+o(t^2)\right\]-2\}\{t^2\}\\\\ &=-\frac\{1\}\{4\}. \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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121、 (3)、 求极限 $\displaystyle \lim\_\{x\to 0\}\frac\{\int\_\{x^2\}^x \tan^3t\mathrm\{ d\} t\}\{\ln (1+x^2)\cdot (\mathrm\{e\}^\{2x^2\}-1)\}$. (吉林大学2023年数学分析考研试题) [函数极限 ]
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https://mp.weixin.qq.com/s/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\_\{x^2\}^x \tan^3t\mathrm\{ d\} t\}\{x^2\cdot 2x^2\} =\frac\{1\}\{2\}\lim\_\{x\to 0\}\frac\{\int\_\{x^2\}^x \tan^3t\mathrm\{ d\} t\}\{x^4\}\\\\ \xlongequal\{\tiny\mbox\{L'Hospital\}\}&\frac\{1\}\{2\}\lim\_\{x\to 0\}\frac\{\tan^3x-\tan^3(x^2)\cdot 2x\}\{4x^3\} \xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\} \frac\{1\}\{2\}\left(\frac\{1\}\{4\}-0\right)=\frac\{1\}\{8\}. \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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122、 1、 计算题. (0-9)、 求 $\displaystyle \lim\_\{x\to 0\}\frac\{\sin x-\tan x\}\{\left(\sqrt[3]\{1+x^2\}-1\right)\left(\sqrt\{1+\sin x\}-1\right)\}$. (吉林师范大学2023年(学科数学)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\} \lim\_\{t\to 0\}\frac\{\sqrt[\alpha]\{1+t\}-1\}\{t\} \xlongequal\{\tiny\mbox\{L'Hospital\}\} \lim\_\{t\to 0\}\alpha(1+t)^\{\frac\{1\}\{\alpha\}-1\}=\alpha \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\{\tan x(\cos x-1)\}\{\frac\{1\}\{3\}x^2\cdot \frac\{1\}\{2\}\sin x\} \xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\} \lim\_\{x\to 0\}\frac\{x\left(-\frac\{x^2\}\{2\}\right)\}\{\frac\{1\}\{6\}x^3\}=-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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123、 (0-10)、 求 $\displaystyle \lim\_\{x\to 1\}x^\frac\{1\}\{1-x\}$. (吉林师范大学2023年(学科数学)数学分析考研试题) [函数极限 ]
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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\}=\mathrm\{exp\}\left\[\lim\_\{x\to 1\}\frac\{1\}\{1-x\}\ln x\right\] \stackrel\{x-1=t\}\{=\}\mathrm\{exp\}\left\[-\lim\_\{t\to 0\}\frac\{1\}\{t\}\ln (1+t)\right\] \xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\} \mathrm\{e\}^\{-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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124、 (2)、 (可能有误) 求极限 $\displaystyle \lim\_\{x\to\infty\}\left(\cos \sqrt[3]\{x+1\}-\cos \sqrt[3]\{x\}\right)x^\frac\{1\}\{3\}$. [不管怎样, 张祖锦还是做出来了] (暨南大学2023年数学分析考研试题) [函数极限 ]
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\begin\{aligned\} \mbox\{原式\}&\xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\} \lim\_\{x\to\infty\}[-\sin \xi\_x] \left(\sqrt[3]\{x+1\}-\sqrt[3]\{x\}\right)x^\frac\{1\}\{3\}\\\\ &=\lim\_\{x\to\infty\} [-\sin \xi\_x] \frac\{x^\frac\{1\}\{3\}\}\{(x+1)^\frac\{2\}\{3\}+x^\frac\{1\}\{3\}(x+1)^\frac\{1\}\{3\}+x^\frac\{2\}\{3\}\} \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\}
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125、 (2)、 求 $\displaystyle \lim\_\{x\to 0\}\frac\{\cos x-\mathrm\{e\}^\{-\frac\{x^2\}\{2\}\}\}\{x^2\tan^2x\}$. (南昌大学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 x-\mathrm\{e\}^\{-\frac\{x^2\}\{2\}\}\}\{x^4\}\\\\ \xlongequal[\tiny\mbox\{展开\}]\{\tiny\mbox\{Taylor\}\}& \lim\_\{x\to 0\}\frac\{\left\[1-\frac\{x^2\}\{2!\}+\frac\{x^4\}\{4!\}+o(x^4)\right\] -\left\[1+\left(-\frac\{x^2\}\{2\}\right)+\frac\{1\}\{2!\}\left(-\frac\{x^2\}\{2\}\right)^2+o(x^4)\right\]\}\{x^4\}\\\\ =&\frac\{1\}\{4!\}-\frac\{1\}\{4\cdot 2\}=-\frac\{1\}\{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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126、 1、 计算题. (1)、 求极限 $\displaystyle \lim\_\{x\to 0\}\frac\{\sqrt\{1+2x\}-\mathrm\{e\}^x+x^2\}\{\arcsin x-\sin x\}$. (南京航空航天大学2023年数学分析考研试题) [函数极限 ]
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https://mp.weixin.qq.com/s/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\{\boxed\{\begin\{array\}\{c\}\left\[1+\frac\{2x\}\{2\}-\frac\{(2x)^2\}\{8\}+\frac\{(2x)^3\}\{16\}+o(x^3)\right\]\\\\ -\left\[1+x+\frac\{x^2\}\{2\}+\frac\{x^3\}\{6\}+o(x^3)\right\]+x^2\end\{array\}\}\}\{\left\[x+\frac\{x^3\}\{6\}+o(x^3)\right\]-\left\[x-\frac\{x^3\}\{6\}+o(x^3)\right\]\} =\frac\{\frac\{1\}\{2\}-\frac\{1\}\{6\}\}\{\frac\{1\}\{3\}\}=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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127、 1、 计算题. (1)、 求极限 $\displaystyle \lim\_\{x\to 0\}(\mathrm\{e\}^x-\sin x)^\frac\{1\}\{x^2\}$. (南京师范大学2023年数学分析考研试题) [函数极限 ]
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https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{原式\}=&\mathrm\{exp\}\left\[\lim\_\{x\to 0\}\frac\{\ln(\mathrm\{e\}^x-\sin x)\}\{x^2\}\right\] \xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\} \mathrm\{exp\}\left\[\lim\_\{x\to 0\}\frac\{\mathrm\{e\}^x-\sin x-1\}\{x^2\}\right\]\\\\ \xlongequal\{\tiny\mbox\{L'Hospital\}\}& \mathrm\{exp\}\left\[\lim\_\{x\to 0\}\frac\{\mathrm\{e\}^x-\cos x\}\{2x\}\right\] \xlongequal\{\tiny\mbox\{L'Hospital\}\} \mathrm\{exp\}\left\[\lim\_\{x\to 0\}\frac\{\mathrm\{e\}^x+\sin x\}\{2\}\right\]=\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)/
---
128、 2、 (20 分) 求极限
\begin\{aligned\} \lim\_\{x\to 0\}\frac\{\cos \sin x-\mathrm\{e\}^\{\cos x-1\}\}\{\tan^2x-\sin^2x\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(南开大学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\} \tan^2x-\sin^2x=&\tan^2x(1-\cos^2x) =\tan^2x(1+\cos x)(1-\cos x)\\\\ &\sim x^2\cdot 2\cdot \frac\{x^2\}\{2\}=x^4, 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\}=&\lim\_\{x\to 0\}\frac\{\cos \sin x-\mathrm\{e\}^\{-2\sin^2\frac\{x\}\{2\}\}\}\{x^4\}\\\\ \xlongequal[\tiny\mbox\{展开\}]\{\tiny\mbox\{Taylor\}\}&\lim\_\{x\to 0\}\frac\{\left\[1-\frac\{\sin^2x\}\{2\}+\frac\{\sin^4x\}\{4!\}+o(x^5)\right\]s -\left\[1-2\sin^2\frac\{x\}\{2\} +\frac\{4\sin^4\frac\{x\}\{2\}\}\{2\}+o(x^5)\right\]\}\{x^4\}\\\\ =&\lim\_\{x\to 0\}\frac\{-\frac\{\sin^2x\}\{2\}+2\sin^2\frac\{x\}\{2\}\}\{x^4\} +\frac\{1\}\{4!\}-2\left(\frac\{1\}\{2\}\right)^4\\\\ =&\lim\_\{x\to 0\}\frac\{-2\sin^2\frac\{x\}\{2\}\cos^2\frac\{x\}\{2\}+2\sin^2\frac\{x\}\{2\}\}\{x^4\} +\frac\{1\}\{4!\}-2\left(\frac\{1\}\{2\}\right)^4\\\\ =&\lim\_\{x\to 0\}\frac\{2\sin^4\frac\{x\}\{2\}\}\{x^4\} +\frac\{1\}\{4!\}-2\left(\frac\{1\}\{2\}\right)^4 =\frac\{1\}\{4!\}=\frac\{1\}\{24\}. \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)/
---
129、 1、 (60 分) 计算题. (1)、 (20 分) 求极限 $\displaystyle \lim\_\{x\to 0\}\frac\{x\sqrt\{1+x^2\}-x\mathrm\{e\}^\{x^2\}\}\{\arcsin x-\sin x\}$. (山东大学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\{x\left\[1+\frac\{x^2\}\{2\}+o(x^2)\right\]-x\left\[1+x^2+o(x^2)\right\]\}\{\left\[x+\frac\{x^3\}\{6\}+o(x^3)\right\]-\left\[x-\frac\{x^3\}\{6\}+o(x^3)\right\]\}\\\\ =&\frac\{\frac\{1\}\{2\}-1\}\{\frac\{1\}\{6\}-\left(-\frac\{1\}\{6\}\right)\}=-\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\}
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)/
---
130、 1、 计算题 (每题 5 分, 共 20 分). (1)、 求极限 $\displaystyle \lim\_\{x\to 0\}\frac\{\mathrm\{e\}^x-(1+2x)^\frac\{1\}\{2\}\}\{\ln(1+x^2)\}$. (陕西师范大学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\}&\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\}
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)/
---
131、 (2)、 用’$\varepsilon-\delta$ 定义证明: $\displaystyle \lim\_\{x\to 2\}\frac\{x^2-4\}\{x^4-6x^2+8\}=\frac\{1\}\{2\}$. (上海大学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) /
\begin\{aligned\} &\forall\ \varepsilon > 0,\exists\ \delta=\min\left\\{\frac\{1\}\{2\},\frac\{\varepsilon\}\{5\}\right\\} > 0,\mathrm\{ s.t.\} \forall\ 0 < |x-2| < \delta,\\\\ &\left|\frac\{x^2-4\}\{x^4-6x^2+8\}-\frac\{1\}\{2\}\right| =\left|\frac\{(x-2)(x+2)\}\{2(x^2-2)\}\right| < \frac\{\delta\cdot\frac\{5\}\{2\}\}\{2\left\[\left(\frac\{3\}\{2\}\right)^2-2\right\]\}=5\delta < \varepsilon. \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)/
---
132、 (2)、 $\displaystyle \lim\_\{x\to 0\}\left(\frac\{2+3\mathrm\{e\}^\{-\frac\{1\}\{x\}\}\}\{1+\mathrm\{e\}^\{-\frac\{4\}\{x\}\}\}-\frac\{|x|\}\{x\}\right)$. (上海大学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) / 由
\begin\{aligned\} &\lim\_\{x\to 0^+\}\left(\frac\{2+3\mathrm\{e\}^\{-\frac\{1\}\{x\}\}\}\{1+\mathrm\{e\}^\{-\frac\{4\}\{x\}\}\}-\frac\{|x|\}\{x\}\right) =\frac\{2+3\cdot 0\}\{1+0\}-1=1,\\\\ &\lim\_\{x\to 0^-\}\left(\frac\{2+3\mathrm\{e\}^\{-\frac\{1\}\{x\}\}\}\{1+\mathrm\{e\}^\{-\frac\{4\}\{x\}\}\}-\frac\{|x|\}\{x\}\right) =\lim\_\{x\to 0^-\}\left(\frac\{2\mathrm\{e\}^\frac\{4\}\{x\}+3\mathrm\{e\}^\frac\{3\}\{x\}\}\{\mathrm\{e\}^\frac\{4\}\{x\}+1\}-\frac\{-x\}\{x\}\right) =\frac\{2\cdot 0+3\cdot 0\}\{0+1\}=1 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知原式 $\displaystyle =1$.跟锦数学微信公众号. [在线资料](
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---
133、 (2)、 已知 $\displaystyle \lim\_\{x\to 0\}\frac\{\sqrt\{1+f(x)\sin x\}-1\}\{\mathrm\{e\}^\{2x\}-1\}=3$. 求 $\displaystyle \lim\_\{x\to 0\}f(x)$. (首都师范大学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\} 3\xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\}\lim\_\{x\to 0\}\frac\{1\}\{2x\}\cdot \frac\{f(x)\sin x\}\{\sqrt\{1+f(x)\sin x\}+1\} =\frac\{1\}\{4\}\lim\_\{x\to 0\}f(x). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \lim\_\{x\to 0\}f(x)=12$.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)/
---
134、 3、 设 $\displaystyle f(x)$ 在 $\displaystyle (0,+\infty)$ 内有一阶导数. (1)、 证明: 对任意 $\displaystyle A > 0$,
\begin\{aligned\} f(x)=\mathrm\{e\}^\{A-x\}f(A)+\mathrm\{e\}^\{-x\}\int\_A^x [f(t)+f'(t)]\mathrm\{e\}^t\mathrm\{ d\} t. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 设 $\displaystyle \lim\_\{x\to+\infty\}[f(x)+f'(x)]=0$, 证明: $\displaystyle \lim\_\{x\to+\infty\}f(x)=0$. (首都师范大学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\} &\mathrm\{e\}^\{-x\}\int\_A^x [f(t)+f'(t)]\mathrm\{e\}^t\mathrm\{ d\} t\xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\} \mathrm\{e\}^\{-x\}[f(t)\mathrm\{e\}^t]'|\_\{t=A\}^\{t=x\}\\\\ =&\mathrm\{e\}^\{-x\}[f(x)\mathrm\{e\}^x-f(A)\mathrm\{e\}^A]=f(x)-\mathrm\{e\}^\{A-x\}f(A). \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\} \lim\_\{x\to+\infty\}f(x)\overset\{\tiny\mbox\{第1步\}\}\{=\}&\lim\_\{x\to+\infty\}\left\\{\frac\{\mathrm\{e\}^A f(A)\}\{\mathrm\{e\}^x\} +\frac\{\displaystyle\int\_A^x [f(t)+f'(t)]\mathrm\{e\}^t\mathrm\{ d\} t\}\{\mathrm\{e\}^x\}\right\\}\\\\ \xlongequal\{\tiny\mbox\{L'Hospital\}\}&0+\lim\_\{x\to+\infty\}\frac\{[f(x)+f'(x)]\mathrm\{e\}^x\}\{\mathrm\{e\}^x\}\xlongequal\{\tiny\mbox\{题设\}\} 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\}
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)/
---
135、 11、 设 $\displaystyle f(x)$ 在 $\displaystyle x=0$ 处连续, $\displaystyle f'(0)=0$, $\displaystyle f‘(0)$ 存在. 证明:
\begin\{aligned\} \lim\_\{x\to 0\}\frac\{f(x)-f\left(\ln (1+x)\right)\}\{x^3\}=\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\}
(首都师范大学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) / 由 Taylor 公式,
\begin\{aligned\} f(x)&=f(0)+f'(0)x+\frac\{f''(0)\}\{2\}x^2+o(x^2) =f(0)+\frac\{f''(0)\}\{2\}x^2+o(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\}
故
\begin\{aligned\} \mbox\{原式\}&=\lim\_\{x\to 0\}\frac\{\left\[f(0)+\frac\{f''(0)\}\{2\}x^2\right\]-\left\[f(0)+\frac\{f''(0)\}\{2\}\ln^2(1+x)+o(x^2)\right\]\}\{x^3\}\\\\ &=\frac\{f''(0)\}\{2\}\lim\_\{x\to 0\}\frac\{x^2-\ln^2(1+x)\}\{x^3\} =\frac\{f''(0)\}\{2\}\lim\_\{x\to 0\}\frac\{x+\ln (1+x)\}\{x\}\cdot\frac\{x-\ln(1+x)\}\{x^2\}\\\\ &=f''(0)\lim\_\{x\to 0\}\frac\{x-\ln(1+x)\}\{x^2\} =f''(0)\lim\_\{x\to 0\}\frac\{x-\left\[x-\frac\{x^2\}\{2\}+o(x^2)\right\]\}\{x^2\}=\frac\{f''(0)\}\{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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---
136、 (3)、 $\displaystyle \lim\_\{x\to 0\}\frac\{\cos \sin x-\cos x\}\{x^3\arctan x\}$. (四川大学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\{-2\sin\frac\{\sin x+x\}\{2\}\sin\frac\{\sin x+x\}\{2\}\}\{x^4\} \xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\} -2\lim\_\{x\to 0\}\frac\{\frac\{\sin x+x\}\{2\}\cdot \frac\{\sin x-x\}\{2\}\}\{x^4\}\\\\ =&-\frac\{1\}\{2\}\lim\_\{x\to 0\}\left(\frac\{\sin x\}\{x\}+1\right)\frac\{\sin x-x\}\{x^3\} \xlongequal[\tiny\mbox\{展开\}]\{\tiny\mbox\{Taylor\}\} -\frac\{1\}\{2\}(1+1)\left(-\frac\{1\}\{6\}\right)=\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\}
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)/
---
137、 (4)、 $\displaystyle \lim\_\{x\to 0\}\frac\{(x+2)^x-2^x\}\{x\sin x\}$. (四川大学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\}\{2^x\}\cdot \frac\{\left(\frac\{x\}\{2\}+1\right)^x-1\}\{x^2\} =\lim\_\{x\to 0\}\frac\{\mathrm\{e\}^\{x\ln \left(1+\frac\{x\}\{2\}\right)\}-1\}\{x^2\}\\\\ \xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\}&\lim\_\{x\to 0\}\frac\{x\ln\left(1+\frac\{x\}\{2\}\right)\}\{x^2\}\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\}
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)/
---
138、 (2)、 $\displaystyle \lim\_\{x\to 0\}\frac\{(5+\sin x)^x-5^x\}\{\tan^2x\}$. (苏州大学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\}\xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\}&\lim\_\{x\to 0\}\frac\{\mathrm\{e\}^\{x\ln(5+\sin x)\}-\mathrm\{e\}^\{x\ln 5\}\}\{x^2\} \xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\} \lim\_\{x\to 0\}\frac\{\mathrm\{e\}^\{\xi\_x\}x\ln\left(1+\frac\{\sin x\}\{5\}\right)\}\{x^2\}\\\\ \xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\}&\lim\_\{x\to 0\}\frac\{\frac\{\sin x\}\{5\}\}\{x\}=\frac\{1\}\{5\}. \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)/
发表于 2023-3-5 08:41:40
