Answer
$-\infty$
Work Step by Step
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.
$\displaystyle \lim_{x\rightarrow 2^{-}}\frac{x^{2}-2x}{x^{2}-4x+4}=-\infty$
