## Calculus: Early Transcendentals (2nd Edition)

The solution is $$\lim_{x\to 2}\frac{x^2-2x}{8-6x+x^2}=-1$$
We will evaluate this limit using L'Hopital's rule. "LR" will stand for "Apply L'Hopital's rule" $$\lim_{x\to 2}\frac{x^2-2x}{8-6x+x^2}=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to 2}\frac{(x^2-2x)'}{(8-6x+x^2)'}=\lim_{x\to 2}\frac{2x-2}{-6+2x}=\left[\frac{2\cdot 2-2}{-6+2\cdot 2}\right] =-1.$$