3

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