95 lines
2.6 KiB
Markdown
95 lines
2.6 KiB
Markdown
## 错题记录
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### 题目 01
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$$\lim_{n \to \infty} \left( \frac{4}{\pi} \arctan \frac{n}{n+1} \right)^n = \, ?$$
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### 错误原因
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实际上应该转换为$\frac{0}{0}$ 型极限处理， 在处理题目的时候没能正确认识到
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### 正确答案
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$$\lim_{n \to \infty} \left( \frac{4}{\pi} \arctan \frac{n}{n+1} \right)^n = e^{-\frac{2}{\pi}}$$
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**解法**：
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$$
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\begin{aligned}
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\lim_{n \to \infty} \left( \frac{4}{\pi} \arctan \frac{n}{n+1} \right)^n
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&= \lim_{n \to \infty}
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\exp\left(
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n \ln\left(\frac{4}{\pi} \arctan \frac{n}{n+1}\right)
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\right) \newline
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&\overset{t=\frac{1}{n}}{=}
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\lim_{t \to 0}
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\exp\left(
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\frac{\ln\left(\frac{4}{\pi} \arctan \frac{1}{1+t}\right)}{t}
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\right) \newline
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&\overset{\text{L'Hospital}}{=} \lim_{t \to 0}
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\exp\left(
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\frac{1}{\frac{4}{\pi} \arctan \frac{1}{1+t}}
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\cdot \frac{4}{\pi} \frac{1}{1 + (\frac{1}{1 + t})^2}
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\cdot - \frac{1}{(1 + t)^2}
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\right) \newline
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&= e^{-\frac{2}{\pi}}
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\end{aligned}
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$$
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故极限为 $e^{-\frac{2}{\pi}}$ 。
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### 知识点
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- $\frac{0}{0}$ 型极限处理方法
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- 指数型极限处理方法
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### 题目 02
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设 $ f(x) $ 在 $ x_0 $ 处可导，且 $ x_n = \sin\frac{1}{n} + \frac{1}{n^2} $。求极限
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$$
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\lim_{n \to \infty} \frac{f(x_0 + \frac{1}{n}) - f(x_0 - x_n)}{\sin\frac{1}{n}}.
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$$
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### 错误原因
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没有意识到等价无穷小替换的问题
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### 正确答案
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$$
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\lim_{n \to \infty} \frac{f(x_0 + \frac{1}{n}) - f(x_0 - x_n)}{\sin\frac{1}{n}} = 2f'(x_0).
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$$
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**解法**：
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因为 $f(x)$ 在 $x_0$ 处可导，利用泰勒展开：
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$$
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\begin{aligned}
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f(x_0 + \tfrac{1}{n}) &= f(x_0) + f'(x_0) \cdot \tfrac{1}{n} + o\left(\tfrac{1}{n}\right), \\
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f(x_0 - x_n) &= f(x_0) - f'(x_0) \cdot x_n + o(x_n).
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\end{aligned}
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$$
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由于 $x_n = \sin\frac{1}{n} + \frac{1}{n^2} = \frac{1}{n} + o\left(\frac{1}{n}\right)$，且 $\sin\frac{1}{n} = \frac{1}{n} + o\left(\frac{1}{n}\right)$，有：
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$$
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\begin{aligned}
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f(x_0 + \tfrac{1}{n}) - f(x_0 - x_n)
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&= \left[f(x_0) + f'(x_0)\tfrac{1}{n} + o(\tfrac{1}{n})\right] - \left[f(x_0) - f'(x_0)x_n + o(x_n)\right] \\
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&= f'(x_0)\left(\tfrac{1}{n} + x_n\right) + o\left(\tfrac{1}{n}\right) \\
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&= f'(x_0)\left(\tfrac{1}{n} + \tfrac{1}{n} + o(\tfrac{1}{n})\right) + o\left(\tfrac{1}{n}\right) \\
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&= 2f'(x_0)\tfrac{1}{n} + o\left(\tfrac{1}{n}\right).
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\end{aligned}
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$$
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因此：
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$$
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\lim_{n \to \infty} \frac{f(x_0 + \frac{1}{n}) - f(x_0 - x_n)}{\sin\frac{1}{n}}
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= \lim_{n \to \infty} \frac{2f'(x_0)\frac{1}{n} + o(\frac{1}{n})}{\frac{1}{n} + o(\frac{1}{n})}
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= 2f'(x_0).
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$$
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### 知识点
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- 可导函数的泰勒展开
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- 等价无穷小替换：$\sin\frac{1}{n} \sim \frac{1}{n}$，$x_n \sim \frac{1}{n}$
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- 导数的定义
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