Calculate the limit $\lim \limits_{x \to 2} \left(\frac{x^2+2x-8}{x^2-2x}\right)$, although $x\neq 2$ I should calculate the Limit $\lim \limits_ {x \to 2} \left(\frac{x^2+2x-8}{x^2-2x}\right)$, although I noticed, that $x\neq 2$ must apply. Is the limit undefined? Otherwise, with which steps should I go on to calculate the limit?

Given $$\lim_{x\rightarrow2}\dfrac{x^2+2x-8}{x^2-2x}=\lim_{x\rightarrow2}\dfrac{\color{red}{(x-2)}(x+4)}{x\color{red}{(x-2)}}=\lim_{x\rightarrow2}\dfrac{x+4}{x}=\lim_{x\rightarrow2}1+\dfrac4x = 3$$
OR
You could also use L'Hopital's rule
$$\lim_{x\rightarrow2}\dfrac{x^2+2x-8}{x^2-2x}=\lim_{x\rightarrow2}\dfrac{2x+2}{2x-2}=\dfrac{2(2)+2}{2(2)-2}=\dfrac{6}{2}=3$$
