## University Calculus: Early Transcendentals (3rd Edition)

$$\lim_{x\to0^-}\frac{x^2-3x+2}{x^3-2x^2}=-\infty$$ $$\lim_{x\to2^+}\frac{x^2-3x+2}{x^3-2x^2}=\frac{1}{4}$$ $$\lim_{x\to2^-}\frac{x^2-3x+2}{x^3-2x^2}=\frac{1}{4}$$ $$\lim_{x\to2}\frac{x^2-3x+2}{x^3-2x^2}=\frac{1}{4}$$ $$\lim_{x\to0}\frac{x^2-3x+2}{x^3-2x^2}=-\infty$$
(a) $$A=\lim_{x\to0^+}\frac{x^2-3x+2}{x^3-2x^2}=\lim_{x\to0^+}\frac{(x^2-2x)-(x-2)}{x^2(x-2)}=\lim_{x\to0^+}\frac{(x-2)(x-1)}{x^2(x-2)}$$ $$A=\lim_{x\to0^+}\frac{x-1}{x^2}$$ As $x\to0^+$: $(x-1)\to-1^+\lt0$ $x^2\to0^+\gt0$ So $\frac{(x-1)}{x^2}\lt0$ as a result. That means $\frac{(x-1)}{x^2}$ will approach $-\infty$. In other words, $$A=-\infty$$ (b) $$B=\lim_{x\to2^+}\frac{x^2-3x+2}{x^3-2x^2}=\lim_{x\to2^+}\frac{x-1}{x^2}$$ (we reuse the result from (a) $$B=\frac{2-1}{2^2}=\frac{1}{4}$$ (c) $$C=\lim_{x\to2^-}\frac{x^2-3x+2}{x^3-2x^2}=\lim_{x\to2^-}\frac{x-1}{x^2}$$ (we reuse the result from (a) $$C=\frac{2-1}{2^2}=\frac{1}{4}$$ (d) Because $\lim_{x\to2^-}\frac{x^2-3x+2}{x^3-2x^2}=\lim_{x\to2^+}\frac{x^2-3x+2}{x^3-2x^2}=\frac{1}{4}$, we conclude that $$\lim_{x\to2}\frac{x^2-3x+2}{x^3-2x^2}=\frac{1}{4}$$ (e) To say anything about the limit as $x\to0$, we need to calculate $\lim_{x\to0^-}\frac{x^2-3x+2}{x^3-2x^2}$ first. Reusing the operation from (a): $$E=\lim_{x\to0^-}\frac{x^2-3x+2}{x^3-2x^2}=\lim_{x\to0^-}\frac{x-1}{x^2}$$ As $x\to0^-$: $(x-1)\to-1^-\lt0$ $x^2\to0^+\gt0$ (because $x^2\ge0$ for all $x$) So $\frac{(x-1)}{x^2}\gt0$ as a result. That means $\frac{(x-1)}{x^2}$ will approach $-\infty$. In other words, $$E=-\infty$$ As the function approaches $-\infty$ as $x\to0$ from both the right and the left, we can conclude that $$\lim_{x\to0}\frac{x^2-3x+2}{x^3-2x^2}=-\infty$$