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

$$\lim_{x\to2^+}\frac{x^2-3x+2}{x^3-4x}=\frac{1}{8}$$ $$\lim_{x\to-2^+}\frac{x^2-3x+2}{x^3-4x}=\infty$$ $$\lim_{x\to0^-}\frac{x^2-3x+2}{x^3-4x}=\infty$$ $$\lim_{x\to1^+}\frac{x^2-3x+2}{x^3-4x}=0$$ $\lim_{x\to0}\frac{x^2-3x+2}{x^3-4x}$ does not exist.
(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.