\(\dfrac{lim}{x\rightarrow2}\dfrac{x^2-3x+3}{x-2}\)
ta có : \(\dfrac{lim}{x\rightarrow2}\dfrac{x^2-3x+3}{ }=1\) và \(\dfrac{lim}{x\rightarrow2}\dfrac{x-2}{ }=0\)
\(\Rightarrow\dfrac{lim}{x\rightarrow2}\dfrac{x^2-3x+3}{x-2}=\infty\)
và \(\dfrac{lim}{x\rightarrow2}\dfrac{x^2-3x+3}{x-2}=+\infty\) khi \(x>2\)
\(\dfrac{lim}{x\rightarrow2}\dfrac{x^2-3x+3}{x-2}=-\infty\) khi \(x< 2\)