请问根号下 n2+1-根号下n2+n 的极限是 怎么算的扫码下载作业帮搜索答疑一搜即得答案解析 查看更多优质解析Sqrt(n^2+1)-Sqrt(n^2+n)=〔(n^2+1)-(n^2+n)〕/〔Sqrt(n^2+1)+Sqrt(n^2+n)〕=(1/n-1)/〔Sqrt(1+1/n^2)+Sqrt(1+1/n)〕n→∞,所有含n的项趋向于0,得极限为-1/2是这样...解析看不懂?免费查看同类题视频解析查看解答}
import 泰勒展开\begin{align} \\&\lim_{n\to\infty}{\sin^2(\pi\sqrt{n^2+n})}\\ =&\frac{1}{2}-\frac12\lim_{n\to\infty}\cos(2\pi\sqrt{n^2+n})\\ =&\frac12\lim_{n\to\infty}\left[1-\cos(2n\pi\sqrt{1+\frac1n})\right]\\ =&\frac12\lim_{n\to\infty}\left[1-\cos(2n\pi(1+\frac{1}{2n}+o(\frac1n))\right]\\ =&\frac12\lim_{n\to\infty}\left[1-\cos(2n\pi+\pi+o(1))\right]\\ =&\frac12\left[1-\cos(\pi+o(1))\right]\\ =&\frac12\left(1-\cos\pi\right)\\ =&\frac12(1+1)\\ =&1 \end{align} 更多泰勒展开}