Tìm các giới hạn sau: a) lim x tién tới -2 của x^2-4/x+2

Giải SBT Toán 11 Bài 2: Giới hạn của hàm số

Bài 3 trang 84 SBT Toán 11 Tập 1: Tìm các giới hạn sau:

a) $\underset{x\to -2}{\lim }\frac{x^2-4}{x+2};$

b) $\underset{x\to 1}{\lim }\frac{x^3-1}{1-x};$

c) $\underset{x\to 3}{\lim }\frac{x^2-4x+3}{x-3};$

d) $\underset{x\to -2}{\lim }\frac{2-\sqrt{x+6}}{x+2};$

e) $\underset{x\to 0}{\lim }\frac{x}{\sqrt{x+1}-1};$

g) $\underset{x\to 2}{\lim }\frac{x^2-4x+4}{x^2-4}.$

Lời giải:

a) $\underset{x\to -2}{\lim }\frac{x^2-4}{x+2}=\underset{x\to -2}{\lim }\frac{\left(x+2\right)\left(x-2\right)}{x+2}$$=\underset{x\to -2}{\lim }\left(x-2\right)=-2-2=-4.$

b) $\underset{x\to 1}{\lim }\frac{x^3-1}{1-x}=\underset{x\to 1}{\lim }\frac{\left(x-1\right)\left(x^2+x+1\right)}{-\left(x-1\right)}$

$=\underset{x\to 1}{\lim }\left.-\left(x^2+x+1\right)\right]=-\left(\underset{x\to 1}{\lim }{x}^2+\underset{x\to 1}{\lim }x+1\right)=-3.$

c) $\underset{x\to 3}{\lim }\frac{x^2-4x+3}{x-3}=\underset{x\to 3}{\lim }\frac{\left(x-1\right)\left(x-3\right)}{x-3}$$=\underset{x\to 3}{\lim }\left(x-1\right)=3-1=2$

d) $\underset{x\to -2}{\text{lim}}\frac{2-\sqrt{x+6}}{x+2}=\underset{x\to -2}{\text{lim}}\frac{\left(2-\sqrt{x+6}\right)\left(2+\sqrt{x+6}\right)}{\left(x+2\right)\left(2+\sqrt{x+6}\right)}$

$=\underset{x\to -2}{\text{lim}}\frac{4-\left(x+6\right)}{\left(x+2\right)\left(2+\sqrt{x+6}\right)}=\underset{x\to -2}{\text{lim}}\frac{-\left(x+2\right)}{\left(x+2\right)\left(2+\sqrt{x+6}\right)}$

$=\underset{x\to -2}{\text{lim}}\frac{-1}{2+\sqrt{x+6}}=\frac{-1}{2+\sqrt{-2+6}}=-\frac{1}{4}.$

e) $\underset{x\to 0}{\text{lim}}\frac{x}{\sqrt{x+1}-1}=\underset{x\to 0}{\text{lim}}\frac{x\left(\sqrt{x+1}+1\right)}{\left(\sqrt{x+1}-1\right)\left(\sqrt{x+1}+1\right)}$

$=\underset{x\to 0}{\text{lim}}\frac{x\left(\sqrt{x+1}+1\right)}{\left(x+1\right)-1}=\underset{x\to 0}{\text{lim}}\left(\sqrt{x+1}+1\right)=2.$

g) $\underset{x\to 2}{\lim }\frac{x^2-4x+4}{x^2-4}=\underset{x\to 2}{\lim }\frac{{\left(x-2\right)}^2}{\left(x+2\right)\left(x-2\right)}$

$=\underset{x\to 2}{\lim }\frac{x-2}{x+2}=\frac{\underset{x\to 2}{\lim }\left(x-2\right)}{\underset{x\to 2}{\lim }\left(x+2\right)}=\frac{0}{4}=0.$