$\lim\limits_{x \to 2^+} \dfrac{\sqrt{3x+1}-2}{x^2-3x+2}$

Vì  $\begin{cases} \lim\limits_{x \to 2^+}\sqrt{3x+1}-2=\sqrt{3.2+1}-2=\sqrt{7}-2>0\\\lim\limits_{x \to 2^+}x^2-3x+2=2^2-3.2+2=0,x^2-3x+2>0 \forall x>2 \end{cases}$

`=>`$\lim\limits_{x \to 2^+} \dfrac{\sqrt{3x+1}-2}{x^2-3x+2}=+\infty$

`=>` `x=2` là tiệm cận đứng

$\lim\limits_{x \to +\infty} \dfrac{\sqrt{3x+1}-2}{x^2-3x+2}$

$=\lim\limits_{x \to +\infty} \dfrac{\sqrt{\dfrac{3}{x^3}+\dfrac{1}{x^4}}-\dfrac{2}{x^2}}{1-\dfrac{3}{x}+\dfrac{2}{x^2}}$

`= (sqrt(0+0)-0)/(1-0+0)`

`=0`

`=> y=0` là tiệm cận ngang