添加slug，改版文章目录命名结构，添加了一个实用的新建文章Shell脚本。
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date = '2025-12-15T16:40:05+08:00'
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draft = false
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title = '2 – 数院人的一天'
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tags = ['数学分析']
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categories = 'math'
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description = '我是数院的，数学再差也是数院的。'
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今日习题：来自《数学分析》上册 习题7.2 L'Hôpital法则。
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## 题目描述1
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求极限：$\lim_{x\to0}{(\frac{a_1^x+a_2^x+\cdots+a_n^x}{n})^\frac{1}{x}}, a_i≥0$
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## 解答
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\[
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\begin{align}
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\lim_{x\to0}{(\frac{a_1^x+a_2^x+\cdots+a_n^x}{n})^\frac{1}{x}}&=\lim_{x\to0}{e^\frac{\ln{(\frac{a_1^x+a_2^x+\cdots+a_n^x}{n})}}{x}}\\
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&=e^{\lim_{x\to0}\frac{\ln{\frac{a_1^x+a_2^x+\cdots+a_n^x}{n}}}{x}}\\
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(L'Hôpital)&=e^{\lim_{x\to0}{\frac{\frac{a_1^x\ln a_1+a_2^x\ln a_2+\cdots+a_n^x\ln a_n^x}{n}}{\frac{a_1^x+a_2^x+\cdots+a_n^x}{n}}}}\\
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&=e^{\frac{\ln a_1+\ln a_2+\cdots+\ln a_n}{n}}\\
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&=\sqrt[n]{a_1a_2\cdots a_n}
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\end{align}
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\]
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## 注
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L'Hôpital习题真难😭
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---
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## 题目描述2
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求极限：$\lim_{x\to0}{\frac{(1+\ln(1+x))^{\frac{1}{\tan x}}+e(x-1)}{x^2}}$
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## 解答
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$$
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\begin{align}
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\lim_{x\to0}{\frac{(1+\ln(1+x))^{\frac{1}{\tan x}}+e(x-1)}{x^2}}&=e\lim_{x\to0}{\frac{e^{\frac{\ln(1+\ln(1+x))}{\tan x}-1}+x-1}{x^2}}\\
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\end{align}
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$$
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由于
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$$
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\begin{align}
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\ln(1+\ln(1+x))&=\ln(1+x-\frac{x^2}{2}+\frac{x^3}{3}+O(x^4))\\
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&=x-\frac{x^2}{2}+\frac{x^3}{3}+O(x^4)-\frac{(x-\frac{x^2}{2}+\frac{x^3}{3}+O(x^4))^2}{2}+\frac{x^3}{3}+O(x^4)\\
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&=x-x^2+\frac{7x^3}{6}+O(x^4)\\
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\tan x&=x+\frac{x^3}{3}+O(x^5)\\
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e^t&=1+t+\frac{t^2}{2}+O(t^3)
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\end{align}
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$$
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则
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$$
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\begin{align}
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\lim_{x\to0}{\frac{(1+\ln(1+x))^{\frac{1}{\tan x}}+e(x-1)}{x^2}}&=e\lim_{x\to0}{\frac{e^{\frac{x-x^2+\frac{7x^3}{6}+O(x^4)}{x+\frac{x^3}{3}+O(x^5)}-1}+x-1}{x^2}}\\
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&=e\lim_{x\to0}{\frac{e^{\frac{-x^2+\frac{5x^3}{6}+O(x^4)}{x+\frac{x^3}{3}+O(x^5)}}+x-1}{x^2}}\\
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&=e\lim_{x\to0}{\frac{e^{\frac{-x+\frac{5x^2}{6}+O(x^3)}{1+\frac{x^2}{3}+O(x^4)}}+x-1}{x^2}}\\
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&=e\lim_{x\to0}{\frac{\frac{-x+\frac{5x^2}{6}+O(x^3)}{1+\frac{x^2}{3}+O(x^4)}+\frac{x^2+O(x^3)}{2(1+\frac{x^2}{3}+O(x^4))^2}+O(x^3)+x}{x^2}}\\
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&=e\lim_{x\to0}{\frac{\frac{\frac{5x^2}{6}+O(x^3)}{1+\frac{x^2}{3}+O(x^4)}+\frac{x^2+O(x^3)}{2(1+\frac{x^2}{3}+O(x^4))^2}+O(x^3)}{x^2}}\\
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&=e\lim_{x\to0}{(\frac{\frac{5}{6}+O(x)}{1+\frac{x^2}{3}+O(x^4)}+\frac{1+O(x)}{2(1+\frac{x^2}{3}+O(x^4))^2}+O(x))}\\
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&=\frac{4e}{3}
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\end{align}
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$$
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## 注
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纯Taylor，不L'Hôpital。
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-----
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## 题目描述3：
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设$f$是方程$f''(x)+3f'(x)+4f(x)=\frac{4x^2}{x^2+x+1}, x≥0, f(0)=1, f'(0)=2$的解。
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证明：$\lim_{x\to+\infty}f''(x)=\lim_{x\to+\infty}f'(x)=0$。
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## 类题：（来自习题课助教黄婧扬）
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(1) 设 \( f(x) \) 在 \( (0, \infty) \) 上可导，\( a > 0 \)。若
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\[
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\lim_{x \to \infty} \big( a f(x) + f'(x) \big) = l,
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\]
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则
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\[
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\lim_{x \to \infty} f(x) = \frac{l}{a}.
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\]
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**证明**：构造辅助函数
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\[
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g(x) = e^{a x} f(x).
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\]
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则
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\[
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g'(x) = e^{a x} \big( a f(x) + f'(x) \big).
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\]
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由题设条件可知
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\[
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\lim_{x \to \infty} \frac{g(x)}{e^{a x}} = \lim_{x \to \infty} \big( a f(x) + f'(x) \big) = l.
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\]
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由 L'Hôpital 法则，我们有
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\[
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\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{g(x)}{e^{a x}} = \lim_{x \to \infty} \frac{g'(x)}{a e^{a x}} = \frac{l}{a}.
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\]
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上述证明对实部为正的复数 \( \alpha \) 都成立。
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(2) 设 \( f(x) \) 在 \( (0, \infty) \) 上二次可导。若有
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\[
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\lim_{x \to \infty} \big( f(x) + f'(x) + f''(x) \big) = l,
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\]
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则
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\[
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\lim_{x \to \infty} f(x) = l.
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\]
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**证明**：设 \( \alpha = \frac{1}{2} - \frac{\sqrt{3}}{2} i \)，\( \beta = \frac{1}{2} + \frac{\sqrt{3}}{2} i \)，则 \( \alpha, \beta \) 为方程 \( x^2 - x + 1 = 0 \) 的两个根。构造辅助函数
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\[
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g(x) = e^{\alpha x} \big(\beta f(x) + f'(x) \big).
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\]
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则
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\[
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g'(x) = e^{\alpha x} \big(\alpha\beta f(x) + (\alpha+\beta)f'(x) + f''(x) \big).\\
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g'(x) = e^{\alpha x} \big(f(x) + f'(x) + f''(x) \big).
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\]
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由题设条件可知
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\[
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\lim_{x \to \infty} \frac{g(x)}{e^{\alpha x}} = \lim_{x \to \infty} \big( f(x) + f'(x) + f''(x) \big) = l.
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\]
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由 L'Hôpital 法则，我们有
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\[
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\lim_{x \to \infty} \big( f(x) + \beta f'(x) \big) = \lim_{x \to \infty} \frac{g'(x)}{\alpha e^{\alpha x}} = \frac{l}{\alpha}.
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\]
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由 (1) 得到
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\[
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\lim_{x \to \infty} f(x) = l.
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\]
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## 解答
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设$g(x)=e^{\alpha x}(\beta f(x)+f'(x))$，待定系数$\alpha,\beta$满足：
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\[
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\begin{align}
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g'(x)&= e^{\alpha x}(\alpha\beta f(x) + (\alpha+\beta)f'(x) + f''(x))\\
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&=e^{\alpha x}(4f(x) +3f'(x) + f''(x))
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\end{align}
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\]
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即：$\alpha,\beta$是方程$x^2-3x+4=0$的两个复根。$\alpha=\frac{3-\sqrt7i}{2},\beta=\frac{3+\sqrt7i}{2}$。
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则：
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\[
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\lim_{x\to+\infty}\frac{g'(x)}{e^{\alpha x}}=\lim_{x\to+\infty}\frac{4x^2}{x^2+x+1}=4\\
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\lim_{x\to+\infty}\frac{g'(x)}{\alpha e^{\alpha x}}=\frac{4}\alpha=\lim_{x\to+\infty}\frac{g(x)}{e^{\alpha x}}=\lim_{x\to+\infty}{(\beta f(x)+f'(x))}
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\]
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由上述类题1的(1)可得：$\lim_{x\to+\infty}{f(x)}=\frac{4}{\alpha\beta}=1$，
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则由$\lim_{x\to+\infty}{(f''(x)+3f'(x)+4f(x))}=\lim_{x\to+\infty}\frac{4x^2}{x^2+x+1}=4$得：
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\[
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\lim_{x\to+\infty}{(f''(x)+3f'(x))}=0
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\]
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由于$\lim_{x\to+\infty}{f(x)}=\frac{4}{\alpha\beta}=1$，
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易知$\lim_{x\to+\infty}{f'(x)}=0$，
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则$\lim_{x\to+\infty}{f''(x)}=0$，原命题得证。
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## 注
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不知道为什么(1)在复数也成立，好神奇。
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