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

$$\lim_{x\to0^+}\Big(\frac{x^2}{2}-\frac{1}{x}\Big)=-\infty$$ $$\lim_{x\to0^-}\Big(\frac{x^2}{2}-\frac{1}{x}\Big)=\infty$$ $$\lim_{x\to\sqrt[3]2}\Big(\frac{x^2}{2}-\frac{1}{x}\Big)=0$$ $$\lim_{x\to-1}\Big(\frac{x^2}{2}-\frac{1}{x}\Big)=\frac{3}{2}$$
$$f(x)=\Big(\frac{x^2}{2}-\frac{1}{x}\Big)=\frac{x^3-2}{2x}$$ (a) As $x\to0^+$, $(2x)\to0^+\gt0$, while $(x^3-2)\to-2^+\lt0$. $\frac{x^3-2}{2x}\lt0$ as a result. That means $\frac{x^3-2}{2x}$ will approach $-\infty$. In other words, $$\lim_{x\to0^+}\Big(\frac{x^2}{2}-\frac{1}{x}\Big)=-\infty$$ (b) As $x\to0^-$, $(2x)\to0^-\lt0$, while $(x^3-2)\to-2^-\lt0$. $\frac{x^3-2}{2x}\gt0$ as a result. That means $\frac{x^3-2}{2x}$ will approach $\infty$. In other words, $$\lim_{x\to0^-}\Big(\frac{x^2}{2}-\frac{1}{x}\Big)=\infty$$ (c) $$\lim_{x\to\sqrt[3]2}f(x)=\lim_{x\to\sqrt[3]2}\frac{x^3-2}{2x}$$ $$\lim_{x\to\sqrt[3]2}f(x)=\frac{(\sqrt[3]2)^3-2}{2\sqrt[3]2}=\frac{2-2}{2\sqrt[3]2}=\frac{0}{2\sqrt[3]2}=0$$ (d) $$\lim_{x\to-1}f(x)=\lim_{x\to-1}\frac{x^3-2}{2x}$$ $$\lim_{x\to-1}f(x)=\frac{(-1)^3-2}{2(-1)}=\frac{-1-2}{-2}=\frac{-3}{-2}=\frac{3}{2}$$