Tính các giới hạn sau:

a)$\underset{x\to -3}{lim}\frac{x-1}{x+1}$

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

c)$\underset{x\to 6}{lim}\frac{\sqrt{x+3}-3}{x-6}$

d)$\underset{x\to +\infty }{lim}\frac{2x-6}{4-x}$

e)$\underset{x\to +\infty }{lim}\frac{17}{x^2+1}$

f)$\underset{x\to +\infty }{lim}\frac{-2x^2+x-1}{3+x}$

Giải: 

a) Ta có 

$\underset{x\to -3}{lim}\frac{x-1}{x+1}=\underset{x\to -3}{lim}(x-1)= -3-1=-4$

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

c)$\underset{x\to 6}{lim}\frac{\sqrt{x+3}-3}{x-6}=\underset{x\to 6}{lim}\frac{x+3-9}{(x-6) (\sqrt{x+3}-3)}=\underset{x\to 6}{lim}\frac{1}{\sqrt{x+3}-3}=\frac{1}{6}$

d)$\underset{x\to +\infty }{lim}\frac{2x-6}{4-x}=\underset{x\to +\infty }{lim}\frac{x\left(2-{\displaystyle \frac{6}{x}}\right)}{x\left({\displaystyle \frac{4}{x}}-1\right)}=\underset{x\to +\infty }{lim}\frac{2-\frac{6}{x}}{\frac{4}{x}-1}=\frac{-2}{1}=-2$

e) Vì $\underset{x\to +\infty }{lim}$$\left(x^2+1\right)=+\infty$nên$\underset{x\to +\infty }{lim}\frac{17}{x^2+1}=0$

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

Vì$\underset{x\to +\infty }{lim}=+\infty và \underset{x\to +\infty }{lim}$$\frac{-2+\frac{1}{x}-\frac{1}{x^2}}{\frac{3}{x}+1}=-2<0$