Question

Tìm giới hạn :

\(lim_{x\rightarrow2^+}\dfrac{x^2-3x+3}{x-2}\)

Answer 1

\(\dfrac{lim}{x\rightarrow2}\dfrac{x^2-3x+3}{x-2}\)

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 \(x< 2\)