一、题目![题目 - 荒原之梦](https://documents.zhaokaifeng.com/uploads/2017/06/06/f68a9e590526998388b0f9b71bd5d3f73dda4ed9764819fe8f36488fa537e9b9499f465fd201d7c117b8901c3ad071915a34a688058a739ebc39835753a8d7cc.svg)
$$
I = \lim \limits_{n \rightarrow \infty}\left\{\left[\sin \left(\frac{\pi}{4}+\frac{1}{n}\right)\right]^{n}+\left[\sin \left(\frac{\pi}{2}+\frac{1}{n}\right)\right]^{n}\right \}=?
$$
难度评级:
二、解析 ![解析 - 荒原之梦](https://documents.zhaokaifeng.com/uploads/2017/06/06/6fff698aa5c66c6c7a143e3d2a00fa8ee7eab76be5360d89eb43a03143848e8cd60377c76bf830c93ec6603be5af661d9c52238834792ea548bf14de10b05ad9.svg)
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:
$$
\lim \limits_{n \rightarrow \infty}\left\{\left[\sin \left(\frac{\pi}{4}+\frac{1}{n}\right)\right]^{n}+\left[\sin \left(\frac{\pi}{2}+\frac{1}{n}\right)\right]^{n}\right\}=0+1=1.
$$
高等数学![箭头 - 荒原之梦](https://documents.zhaokaifeng.com/uploads/2017/06/06/c19692009799eac2a7eb5b9d73167ae3dd6cad169ea3ccdbeb97491b80e87593cfa7384844ec1720d0fb9cf5f00ac456f249d047b61ce2d90bdd241e042f4d89.svg)
涵盖高等数学基础概念、解题技巧等内容,图文并茂,计算过程清晰严谨。
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通过专题的形式对数学知识结构做必要的补充,使所学知识更加连贯坚实。