Description
证明下列关系式:
(1). \(\arcsin x=x+o(x),x\to 0\);\(\quad\)(2). \(\arctan x=x+o(x)\);\(\quad\)(3). \(\sqrt[n]{1+x}=1+\dfrac{1}{n}x+o(x),x\to 0\);
(4). \(\sqrt{1+\tan x}-\sqrt{1+\sin x}\sim\dfrac{1}{4}x^3,x\to0\);\(\quad\)(5). \(\sqrt{x+\sqrt{1+\sqrt{x}}}\sim\sqrt{x},x\to+\infty\);\(\quad\)(6). \(1+\cos\pi x\sim\dfrac{\pi^2}{2}(x-1)^2,x\to 1\).
Solution
(1).
已知 \(x\to 0\) 时有 \(\sin x\sim x\), 令 \(y=\sin x\), 则 \(y\to 0\), 且 \(\arcsin y=x\sim y\). 即当 \(x\to 0\) 时, \(\arcsin x\) 和 \(x\) 是等价无穷小, 于是由书中定理 4.4 即得 \(\arcsin x = x+o(x)\).
(2).
已知 \(x\to 0\) 时有 \(\tan x\sim x\), 与 (1) 类似地有 \(\arctan x\sim x\), 于是有 \(\arctan x=x+o(x)\).
(3).
\[\sqrt[n]{1+x}-1=\dfrac{x}{(1+x)^\frac{n-1}{n}+(1+x)^{\frac{n-2}{n}}+\cdots+1} \]于是有
\[\lim_{x\to 0}\sqrt[n]{1+x}-1=\dfrac{1}{n}x \]即
\[\sqrt{1+x}= \dfrac{1}{n}x+1+o(x) \](4).
\[\begin{aligned} \lim_{x\to 0}\sqrt{1+\tan x}-\sqrt{1+\sin x}&=\lim_{x\to 0}1+\dfrac{1}{2}\tan x-1-\dfrac{1}{2}\sin x+o(\tan x)+o(\sin x)\\ &=\lim_{x\to 0}\dfrac{1}{2}(\tan x-\sin x)+o(\tan x)+o(\sin x)\\ &=\lim_{x\to 0}\dfrac{1}{2}\sin x\left(\dfrac{1}{\cos x}-1\right)+o(\tan x)+o(\sin x)\\ &=\lim_{x\to 0}\dfrac{1}{2}x(1-\cos x)+o(x)\\ &=\dfrac{1}{4}x^3+o(x) \end{aligned} \]因而 \(\sqrt{1+\tan x}-\sqrt{1+\sin x}\sim \dfrac{1}{4}x^3,x\to 0\).
(5).
\[\lim_{x\to+\infty}\dfrac{\sqrt{x+\sqrt{1+\sqrt{x}}}}{\sqrt{x}}=\sqrt{1+\sqrt{\dfrac{1+\sqrt x}{x}}}=\sqrt{1+x^{-\frac{1}{4}}}=1 \]因此 \(\sqrt{x+\sqrt{1+\sqrt{x}}}\sim\sqrt{x},x\to+\infty\).
(6).令 \(t=x-1\), 则原式为
\[1+\cos(\pi t-\pi)=1-\cos \pi t\sim \dfrac{\pi^2}{2}t^2,t\to 0 \] 标签：1.4,dfrac,sqrt,Chap.1,tan,lim,习题,sin,sim From： https://www.cnblogs.com/hl-fc/p/16793465.html