Question

tìm giới hạn sau:

\(\lim\limits_{x\rightarrow3}\) \(\dfrac{\sqrt{x^2-3x+9}-2x+3}{2x-6}\)

Answer 1

\(f\left(x\right)=\dfrac{\sqrt{x^2-3x+9}-\left(2x-3\right)}{2x-6}\)

\(=\dfrac{x^2-3x+9-4x^2+12x-9}{2\left(x-3\right)\left(\sqrt{x^2-3x+9}+\left(2x-3\right)\right)}\)

\(=\dfrac{-3x^2+9x}{2\left(x-3\right)\left(\sqrt{x^2-3x+9}+\left(2x-3\right)\right)}\)

\(=\dfrac{-3x\left(x-3\right)}{2\left(x-3\right)\left(\sqrt{x^2-3x+9}+\left(2x-3\right)\right)}\)

\(=\dfrac{-3x}{2\left(\sqrt{x^2-3x+9}+\left(2x-3\right)\right)}\)

\(\lim\limits_{x\rightarrow3}f\left(x\right)=\dfrac{-3.3}{2\left(\sqrt{9-9+9}+\left(6-3\right)\right)}=-\dfrac{3}{4}\)