# 这道三角函数极限题你能秒解吗

## 一、题目

$$I = \lim \limits_{n \rightarrow \infty}\left\{\left[\sin \left(\frac{\pi}{4}+\frac{1}{n}\right)\right]^{n}+\left[\sin \left(\frac{\pi}{2}+\frac{1}{n}\right)\right]^{n}\right \}=?$$

## 二、解析

### 秒解法

$$\lim \limits_{n \rightarrow \infty}\left\{\left[\sin \left(\frac{\pi}{4}+\frac{1}{n}\right)\right]^{n}+\left[\sin \left(\frac{\pi}{2}+\frac{1}{n}\right)\right]^{n}\right\}.$$

$$\lim \limits_{n \rightarrow \infty} \sin \left(\frac{\pi}{4}+\frac{1}{n}\right)=\lim \limits_{n \rightarrow \infty} \sin \left(\frac{\pi}{4}+0\right)=\sin \frac{\pi}{4}=\frac{\sqrt{2}}{2}<1$$

$$\lim \limits_{n \rightarrow \infty} \sin \left(\frac{\pi}{2}+\frac{1}{n}\right)=\lim \limits_{n \rightarrow \infty} \sin \left(\frac{\pi}{2}+0\right)=\sin \frac{\pi}{2}=1$$

$$I=\lim \limits_{n \rightarrow \infty}\left(\frac{\sqrt{2}}{2}\right)^{n}+\lim \limits_{n \rightarrow \infty} 1^{n}=0+1=1$$

### 严谨解法

$$\frac{\pi}{4}<\frac{\pi}{4}+\frac{1}{n}<\frac{\pi}{3}$$

$$\lim \limits_{n \rightarrow \infty}\left[\sin \frac{\pi}{4}\right]^{n}<\lim \limits_{n \rightarrow \infty}\left[\sin \left(\frac{\pi}{4}+\frac{1}{n}\right)\right]^{n}<\lim \limits_{n \rightarrow \infty}\left[\sin \frac{\pi}{3}\right]^{n} \Rightarrow$$

$$\lim \limits_{n \rightarrow \infty}\left[\frac{\sqrt{2}}{2}\right]^{n}<\lim \limits_{n \rightarrow \infty}\left[\sin \left(\frac{\pi}{4}+\frac{1}{n}\right)\right]^{n}<\lim \limits_{n \rightarrow \infty}\left[\frac{\sqrt{3}}{2}\right]^{n} \Rightarrow$$

$$0<\lim \limits_{n \rightarrow \infty}\left[\sin \left(\frac{\pi}{4}+\frac{1}{n}\right)\right]^{n}<0 \Rightarrow$$

$$\lim \limits_{n \rightarrow \infty} \sin \left(\frac{\pi}{4}+\frac{1}{n}\right)=0.$$

$$\lim \limits_{n \rightarrow \infty}\left[\sin \left(\frac{\pi}{2}+\frac{1}{n}\right)\right]^{n}=\lim \limits_{n \rightarrow \infty}\left(\cos \frac{1}{n}\right)^{n}=$$

$$\lim \limits_{n \rightarrow \infty}\left(\cos \frac{1}{n}-1+1\right)^{\frac{1}{\cos \frac{1}{n}-1} \cdot n\left(\cos \frac{1}{n}-1\right)}=$$

$$e^{\lim \limits_{n \rightarrow \infty} n \left(\cos \frac{1}{n}-1\right)}$$

$$\lim \limits_{n \rightarrow \infty} n\left(\cos \frac{1}{n}-1\right)$$

$$\lim \limits_{n \rightarrow \infty} \frac{\cos \frac{1}{n}-1}{\frac{1}{n}}=\lim \limits_{n \rightarrow \infty} \frac{-\frac{1}{2} \cdot \frac{1}{n^{2}}}{\frac{1}{n}}=\frac{-1}{2 n}=0$$

$$\lim \limits_{n \rightarrow \infty}\left(\cos \frac{1}{n}\right)^{n}=e^{0}$$

$$\lim \limits_{n \rightarrow \infty}\left\{\left[\sin \left(\frac{\pi}{4}+\frac{1}{n}\right)\right]^{n}+\left[\sin \left(\frac{\pi}{2}+\frac{1}{n}\right)\right]^{n}\right\}=0+1=1.$$