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芝士点¶
等价无穷小:等价无穷小的定义:设当\(x\rightarrow x_0\) 时, \(f(x)\)和\(g(x)\)均为无穷小量。若\(\displaystyle\lim_{x\rightarrow x_0}\frac{f(x)}{g(x)}=1\),则称\(f\)和\(g\)是等价无穷小量,记作\(f(x)\)~\(g(x)(x\to x_0)\)。
\(x\to 0\)时常用的等价无穷小:
两个重要极限¶
\({\displaystyle\lim_{x \rightarrow 0}\frac{\sin x}{x} = 1}\)
\({\displaystyle\lim_{x \rightarrow + \infty}(1 + \frac{1}{x})^x = e}\)
你已经学会了等价无穷小,试试看!
例1¶
求\(\displaystyle\lim_{x\to0^+}\frac{1-\sqrt{cosx}}{x(1-cos\sqrt{x})}\)
解:原式=\(\displaystyle\lim_{x\to0^+}\frac{1-cosx}{x(1-cos\sqrt{x})(1+cos\sqrt{x})}=\lim_{x\to0^+}\frac{\frac{1}{2}x^2}{x・\frac{1}{2}x・(1+cos\sqrt{x})}=\frac{1}{2}\)
例2¶
计算\(\displaystyle\lim_{x\to0}\frac {\sqrt{cosx}-\sqrt[3]{cosx}}{sin^2x}\)
我认为的最简解:
\(\because\sqrt{cosx}=1-\frac{x^2}{4}+o(x^2)\)
\(\sqrt[3]{cosx}=1-\frac{x^2}{6}+o(x^2)\)
\(\therefore原式=\displaystyle\lim_{x\to0}\frac {-\frac{1}{12}x^2+o(x^2)}{sin^2x}=\lim_{x\to0}\frac {-\frac{1}{12}x^2+o(x^2)}{x^2}=-\frac{1}{12}\)
例3¶
\(求e^{\displaystyle\lim_{n\to\infty}nln(1+sin\pi\sqrt{1+4n^2})}\)
解:
改写数列表达式:\(\displaystyle sin\pi\sqrt{1+4n^2}=sin(\pi\sqrt{1+4n^2}-2n\pi)=sin\frac{1}{\sqrt{1+4n^2}+2n}\)
\(原式=e^{\displaystyle\lim_{n\to\infty}nln(1+sin\frac{1}{\sqrt{1+4n^2}+2n})}\),其中\(\displaystyle\lim_{n\to\infty}nln(1+sin\frac{1}{\sqrt{1+4n^2}+2n})\\=\displaystyle\lim_{n\to\infty}nsin\frac{1}{\sqrt{1+4n^2}+2n}\\=\lim_{n\to\infty}\frac{n}{\sqrt{1+4n^2}+2n}=\frac{1}{4}\)
\(\therefore\) 原极限=\(e^{\frac{1}{4}}\)
例4¶
求\(\displaystyle \lim_{x\to1}\displaystyle\frac{1-\sqrt[n]{cos2n\pi x}}{(x-1)(x^x-1)}\)
解:略