Tìm các giới hạn sau:

Giải:

a) $\underset{x\to 2}{lim}\frac{x+3}{x^2+x+4}= \frac{2+3}{4+2+4}=\frac{5}{10}=\frac{1}{2}$

b) $\underset{x\to 3}{lim}\frac{x^2+5x +6}{x^{2 }+3}=\underset{x\to 3}{lim}\frac{\left(x+2\right)\left(x+3\right)}{x\left(x+3\right)}= \underset{x\to 3}{lim}\frac{x+2}{x}=\frac{1}{3}$

c) $\underset{x\to 4^{-}}{lim}\frac{2x-5}{x-4}= -\infty$

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

e)$\underset{x\to \infty }{lim}\frac{x+3}{3x-1}=\underset{x\to \infty }{lim}\frac{1+{\displaystyle \frac{3}{x}}}{3-{\displaystyle \frac{1}{x}}}=\frac{1}{3}$

f$\underset{x\to \infty }{lim}\frac{\sqrt{x^2-2x+4}-x}{3x-1}=\underset{x\to \infty }{lim}\frac{-x\sqrt{1-{\displaystyle \frac{2}{x}}+{\displaystyle \frac{4}{x^2}} }-x}{3x-1}= \underset{x\to \infty }{lim}\frac{-\sqrt{1-\frac{2}{x}+\frac{4}{x^2}}-1}{3-{\displaystyle \frac{1}{x}}}=-\frac{2}{3}$