这道三角函数极限题你能秒解吗

一、题目

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

难度评级：

二、解析

秒解法

已知：

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

又：

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

且：

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

则：

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

严谨解法

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

当 $n>4$ 时，有：

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

于是：

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

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

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

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

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

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

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

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

又：

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

于是：

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

即：

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

综上可知：

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

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