44 lines
1.1 KiB
Markdown
44 lines
1.1 KiB
Markdown
## 错题记录
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### 题目
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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 |