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+## 错题记录
+
+### 题目
+$$\lim_{n \to \infty} \left( \frac{4}{\pi} \arctan \frac{n}{n+1} \right)^n = \, ?$$
+
+
+### 错误原因
+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
+
+### 正确答案
+$$\lim_{n \to \infty} \left( \frac{4}{\pi} \arctan \frac{n}{n+1} \right)^n = e^{-\frac{2}{\pi}}$$
+
+**解法**：
+$$
+\begin{aligned}
+\lim_{n \to \infty} \left( \frac{4}{\pi} \arctan \frac{n}{n+1} \right)^n 
+&= \lim_{n \to \infty} 
+\exp\left( 
+    n \ln\left(\frac{4}{\pi} \arctan \frac{n}{n+1}\right) 
+    \right) \newline
+
+&\overset{t=\frac{1}{n}}{=} 
+\lim_{t \to 0} 
+\exp\left( 
+    \frac{\ln\left(\frac{4}{\pi} \arctan \frac{1}{1+t}\right)}{t} 
+    \right) \newline
+
+&\overset{\text{L'Hospital}}{=} \lim_{t \to 0} 
+\exp\left( 
+    \frac{1}{\frac{4}{\pi} \arctan \frac{1}{1+t}} 
+    \cdot \frac{4}{\pi} \frac{1}{1 + (\frac{1}{1 + t})^2}  
+    \cdot (- \frac{1}{1 + t})^2
+    \right) \newline
+
+&= e^{-\frac{2}{\pi}}
+
+\end{aligned}
+$$
+
+故极限为$e^{-\frac{2}{\pi}}$。
+
+### 知识点
+ - $\frac{0}{0}$ 型极限处理方法
+ - 指数型极限处理方法
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