Answer
$$\lim_{x\to2^+}\frac{x^2-3x+2}{x^3-4x}=\frac{1}{8}$$
$$\lim_{x\to-2^+}\frac{x^2-3x+2}{x^3-4x}=\infty$$
$$\lim_{x\to0^-}\frac{x^2-3x+2}{x^3-4x}=\infty$$
$$\lim_{x\to1^+}\frac{x^2-3x+2}{x^3-4x}=0$$
$\lim_{x\to0}\frac{x^2-3x+2}{x^3-4x}$ does not exist.
Work Step by Step
(a) $$A=\lim_{x\to2^+}\frac{x^2-3x+2}{x^3-4x}=\lim_{x\to2^+}\frac{(x^2-2x)-(x-2)}{x(x^2-4)}=\lim_{x\to2^+}\frac{(x-2)(x-1)}{x(x-2)(x+2)}$$ $$A=\lim_{x\to2^+}\frac{x-1}{x(x+2)}$$ $$A=\frac{2-1}{2(2+2)}=\frac{1}{2\times4}=\frac{1}{8}$$
(b) $$B=\lim_{x\to-2^+}\frac{x^2-3x+2}{x^3-4x}$$
Reusing the operation from (a):
$$B=\lim_{x\to-2^+}\frac{x-1}{x(x+2)}$$
As $x\to-2^+\lt0$:
$(x+2)\to0^+\gt0$
$(x-1)\to-3^+\lt0$
So $\frac{x-1}{x(x+2)}\gt0$ as a result.
That means $\frac{x-1}{x(x+2)}$ will approach $\infty$. In other words, $$B=\infty$$
(c) $$C=\lim_{x\to0^-}\frac{x^2-3x+2}{x^3-4x}$$
Reusing the operation from (a):
$$C=\lim_{x\to0^-}\frac{x-1}{x(x+2)}$$
As $x\to0^-\lt0$:
$(x+2)\to2^-\gt0$
$(x-1)\to-1^-\lt0$
So $\frac{x-1}{x(x+2)}\gt0$ as a result.
That means $\frac{x-1}{x(x+2)}$ will approach $\infty$. In other words, $$C=\infty$$
(d) $$D=\lim_{x\to1^+}\frac{x^2-3x+2}{x^3-4x}$$
Reusing the operation from (a):
$$D=\lim_{x\to1^+}\frac{x-1}{x(x+2)}$$ $$D=\frac{1-1}{1\times(1+2)}=\frac{0}{1\times3}=0$$
(e) To say anything about the limit as $x\to0$, we need to calculate $\lim_{x\to0^+}\frac{x^2-3x+2}{x^3-4x}$ first.
Reusing the operation from (a): $$E=\lim_{x\to0^+}\frac{x^2-3x+2}{x^3-4x}=\lim_{x\to0^+}\frac{x-1}{x(x+2)}$$
As $x\to0^+\gt0$:
$(x+2)\to2^+\gt0$
$(x-1)\to-1^+\lt0$
So $\frac{x-1}{x(x+2)}\lt0$ as a result.
That means $\frac{x-1}{x(x+2)}$ will approach $-\infty$. In other words, $$E=-\infty$$
As the function approaches $-\infty$ as $x\to0^+$ and approaches $\infty$ as $x\to0^-$, we can conclude that $\lim_{x\to0}\frac{x^2-3x+2}{x^3-4x}$ does not exist, as the function approaches different values of $x\to0$ from the left and the right.
