Answer
a. $-\infty$
b. $\frac{1}{4}$
c. $\frac{1}{4}$
d. $\frac{1}{4}$
e. $-\infty$
Work Step by Step
a. We have $x\to0^+,x\ne0$, $\lim_{x\to0^+}\frac{x^2-3x+2}{x^3-2x^2}=\lim_{x\to0^+}\frac{x^2-3x+2}{x^2(x-2)}=\lim_{x\to0^+}\frac{1}{x^2}\frac{0^2-0+2}{0-2}=-\lim_{x\to0^+}\frac{1}{x^2}=-\infty$
b. We have $x\to2^+,x\ne2$, $\lim_{x\to2^+}\frac{x^2-3x+2}{x^3-2x^2}=\lim_{x\to2^+}\frac{(x-1)(x-2)}{x^2(x-2)}=\lim_{x\to2^+}\frac{x-1}{x^2}=\frac{2-1}{2^2}=\frac{1}{4}$
c. We have $x\to2^-,x\ne2$, $\lim_{x\to2^-}\frac{x^2-3x+2}{x^3-2x^2}=\lim_{x\to2^-}\frac{(x-1)(x-2)}{x^2(x-2)}=\lim_{x\to2^-}\frac{x-1}{x^2}=\frac{2-1}{2^2}=\frac{1}{4}$
d. Since $\lim_{x\to2^+}\frac{x^2-3x+2}{x^3-2x^2}=\lim_{x\to2^-}\frac{x^2-3x+2}{x^3-2x^2}$, we have $\lim_{x\to2}\frac{x^2-3x+2}{x^3-2x^2}=\frac{1}{4}$
e. We have $x\to0, x\ne0$, $\lim_{x\to0}\frac{x^2-3x+2}{x^3-2x^2}=\lim_{x\to0}\frac{x^2-3x+2}{x^2(x-2)}=\lim_{x\to0}\frac{1}{x^2}\frac{0^2-0+2}{0-2}=-\lim_{x\to0}\frac{1}{x^2}=-\infty$
