Calculus 8th Edition

$-\infty$
Factoring, numerator : $x^{2}-2x=x(x-2)$ denominator: square of a difference $(x-2)^{2}=x^{2}-2\cdot 2x+2^{2}$ So $\displaystyle \lim_{x\rightarrow 2^{-}}\frac{x^{2}-2x}{x^{2}-4x+4}=\lim_{x\rightarrow 2^{-}}\frac{x(x-2)}{(x-2)^{2}}=\lim_{x\rightarrow 2^{-}}\frac{x}{x-2}$ As $x\rightarrow 2^{-}$, the values of x are positive, slightly less than 2. So the numerator is positive, close to 2 and the denominator is negative, approaching 0. $\displaystyle \lim_{x\rightarrow 2^{-}}\frac{x^{2}-2x}{x^{2}-4x+4}=-\infty$