Answer
0
Work Step by Step
$\lim\limits_{x \to 2}\frac{x^2-4x+4}{x^4-3x^2-4}=\lim\limits_{x \to 2}\frac{(x-2)(x-2)}{(x^2-4)(x^2+1)}=\lim\limits_{x \to 2}\frac{(x-2)(x-2)}{(x+2)(x-2)(x^2+1)}=\lim\limits_{x \to 2}\frac{x-2}{(x+2)(x^2+1)}=\frac{(2)-2}{((2)+2)((2)^2+1)}=\frac{0}{20}=0$
