2020-11-07

2020-11-07 本文已影响0人寻松点点

练习题

1. $\lim\limits_{x \to 1} \frac{x^2-3x+2}{x^2+2x-3}$

2. $\lim\limits_{x \to 4}\frac{x-4}{\sqrt{x+5}-3}$

3. $\lim\limits_{x \to 0} \frac{\sqrt[3]{1+x}-\sqrt[3]{1-x}} {x }$

4. $\lim\limits_{x \to 0}\frac{1-\sqrt{1+3x}}{x^2+sin2x}$

解答过程

$\lim_{x \to 1} \frac{x^2-3x+2}{x^2+2x-3} \\ =\lim_{x \to 1} \frac{(x-1)(x-2)}{(x-1)(x+3)} \\ =\lim_{x \to 1}\frac{x-2}{x+3} \\ =- \frac{1}{4}$

$\lim_{x \to 4}\frac{x-4}{\sqrt{x+5}-3} \\ =\lim_{x\to 4}\frac{(x-4)(\sqrt{x+5}+3)}{x+5-9} \quad 分母有理化 \\ =\lim_{x \to 4}\frac{(x-4)(\sqrt{x+5}+3)}{x-4} \\ =\lim_{x \to 4} \sqrt{x+5}+3 \\ =6$

$\lim_{x \to 0}\frac{\sqrt[3]{1+x}-\sqrt[3]{1-x}}{x} \\ =\lim_{x \to 0}\frac{1 }{x } \cdot\frac{(1+x)-(1-x)}{(\sqrt[3]{1+x})^2 +\sqrt[3]{1+x}\sqrt[3]{1-x}+(\sqrt[3]{1-x})^2 } \\ =\lim_{x \to 0} \frac{1}{x}\cdot\frac{2x}{(\sqrt[3]{1+x})^2 +\sqrt[3]{1+x}\sqrt[3]{1-x}+(\sqrt[3]{1-x})^2 } \\ =\frac{2}{3}$

$\lim_{x \to 0}\frac{1-\sqrt{1+3x}}{x^2+sin2x} \\ =\lim_{x \to 0}\frac{1-(1+3x) }{ (x^2+sin2x)\cdot(1+\sqrt{1+3x})} \\ =\lim_{x \to 0}\frac{-3}{1+\sqrt{1+3x} }\cdot\frac{x}{x^2+sin2x} \\ =-\frac{3}{2}\lim_{x \to 0}\frac{1}{x+\frac{sin2x}{x}}= -\frac{3}{2}\lim_{x \to 0}\frac{1}{x+2\cdot \frac{sin2x}{2x}}\\ = -\frac{3}{2} \cdot \frac{1}{2}\\ =-\frac{3}{4}$