Answer
a. $\frac{1}{8} $
b. $\infty$
c. $\infty $
d. $0$
e. Does not exist.
Work Step by Step
a. We have $x\to2^+, x\ne2$ and $\lim_{x\to2^+}\frac{x^2-3x^2+2}{x^3-4x}=\lim_{x\to2^+}\frac{(x-1)(x-2)}{x(x+2)(x-2)}=\lim_{x\to2^+}\frac{x-1}{x(x+2)}=\frac{2-1}{2(2+2)}=\frac{1}{8} $
b. We have $x\to-2^+, x+2\gt0$ and $\lim_{x\to-2^+}\frac{x^2-3x^2+2}{x^3-4x}=\lim_{x\to-2^+}\frac{(x-1)(x-2)}{x(x+2)(x-2)}=\lim_{x\to-2^+}\frac{1}{x+2}\frac{x-1}{x}=\lim_{x\to-2^+}\frac{1}{x+2}\frac{-2-1}{-2}=\infty$
c. We have $x\to0^-, x\lt0$ and $\lim_{x\to0^-}\frac{x^2-3x^2+2}{x^3-4x}=\lim_{x\to0^-}\frac{1}{x}\frac{x-1}{x+2}=\lim_{x\to0^-}\frac{1}{x}\frac{0-1}{0+2}=\infty $
d. We have $x\to1^+, x-1\gt0$ and $\lim_{x\to1^+}\frac{x^2-3x^2+2}{x^3-4x}=\lim_{x\to1^+}\frac{x-1}{x(x+2)}=\frac{1-1}{1(1+2)}=0$
e. We know from part-c that $\lim_{x\to0^-}\frac{x^2-3x^2+2}{x^3-4x}=\infty $
For $x\to0^+, x\gt0$, we have $\lim_{x\to0^+}\frac{x^2-3x^2+2}{x^3-4x}=\lim_{x\to0^+}\frac{1}{x}\frac{x-1}{x+2}=\lim_{x\to0^+}\frac{1}{x}\frac{0-1}{0+2}=-\infty $. Since the left and right hand limits are not equal, the limit of the function at $x\to0$ does not exist.
