Answer
(a) The limit does not exist.
(b) $\lim_{x\to2}\frac{x^2-4x+4}{x^3+5x^2-14x}=0$
Work Step by Step
$$f(x)=\frac{x^2-4x+4}{x^3+5x^2-14x}$$
a) $$\lim_{x\to0}\frac{x^2-4x+4}{x^3+5x^2-14x}=\lim_{x\to0}\frac{(x-2)^2}{x(x-2)(x+7)}=\lim_{x\to0}\frac{x-2}{x(x+7)}$$
As $x\to0$, we have $x(x+7)$ approaches $0$ while $x-2$ approaches $-2$. Therefore, function $\frac{x-2}{x(x+7)}$ will approach $-\infty$, which means the limit does not exist.
$$\lim_{x\to0}\frac{x^2-4x+4}{x^3+5x^2-14x}=-\infty$$
b) $$\lim_{x\to2}\frac{x^2-4x+4}{x^3+5x^2-14x}=\lim_{x\to2}\frac{(x-2)^2}{x(x-2)(x+7)}=\lim_{x\to2}\frac{x-2}{x(x+7)}$$
$$\lim_{x\to2}\frac{x^2-4x+4}{x^3+5x^2-14x}=\frac{2-2}{2(2+7)}=\frac{0}{2\times9}=0$$
