Ngân hàng bài tập
S
Tính các giới hạn sau:
- $\lim\limits_{x\rightarrow-2}\dfrac{x^2+3x+2}{x+2}$.
- $\lim\limits_{x\rightarrow1}\dfrac{\sqrt{x+3}-2x}{x-1}$.
1 lời giải
- $\begin{aligned}[t]\lim\limits_{x\rightarrow-2}\dfrac{x^2+3x+2}{x+2}&=\lim\limits_{x\rightarrow-2}\dfrac{(x+1)(x+2)}{x+2}\\ &=\lim\limits_{x\rightarrow-2}(x+1)=-2+1=-1.\end{aligned}$
- $\begin{aligned}[t]\lim\limits_{x\rightarrow1}\dfrac{\sqrt{x+3}-2}{x-1}&=\lim\limits_{x\rightarrow1}\dfrac{\left(\sqrt{x+3}-2\right)\left(\sqrt{x+3}+2\right)}{(x-1)\left(\sqrt{x+3}+2\right)}\\ &=\lim\limits_{x\rightarrow1}\dfrac{(x+3)-4}{(x-1)\left(\sqrt{x+3}+2\right)}\\ &=\lim\limits_{x\rightarrow1}\dfrac{x-1}{(x-1)\left(\sqrt{x+3}+2\right)}\\ &=\lim\limits_{x\rightarrow1}\dfrac{1}{\sqrt{x+3}+2}=\dfrac{1}{4}.\end{aligned}$
Do đó $\begin{aligned}[t]\lim\limits_{x\rightarrow1}\dfrac{\sqrt{x+3}-2x}{x-1}&=\lim\limits_{x\rightarrow1}\dfrac{\sqrt{x+3}-2-(2x-2)}{x-1}\\ &=\lim\limits_{x\rightarrow1}\dfrac{\sqrt{x+3}-2}{x-1}-\lim\limits_{x\rightarrow1}\dfrac{2x-2}{x-1}\\ &=\dfrac{1}{4}-\lim\limits_{x\rightarrow1}2=-\dfrac{3}{4}.\end{aligned}$