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

The graph is shown below. (a) As $x\to0^+$, the graph approaches $\infty$. (b) As $x\to\pm\infty$, the graph approaches $\infty$. (c) As $x\to1$, the graph approaches $0$ and as $x\to-1$, the graph also approaches $0$.
$$y=\frac{3}{2}\Big(x-\frac{1}{x}\Big)^{2/3}$$ The graph is shown below. (a) As $x\to0^+$, the graph approaches $\infty$. Reason: As $x\to0^+$, $$1/x\to\infty$$ $$x-\frac{1}{x}\to-\infty$$ $$\frac{3}{2}\Big(x-\frac{1}{x}\Big)^{2/3}\to\infty$$ ($A^{2/3}$ is always positive) (b) As $x\to\pm\infty$, the graph approaches $\infty$. Reason: As $x\to\pm\infty$, $$1/x\to0$$ $$\frac{3}{2}\Big(x-\frac{1}{x}\Big)^{2/3}\to\infty$$ ($A^{2/3}$ is always positive) (c) As $x\to1$, the graph approaches $0$ and as $x\to-1$, the graph also approaches $0$. Reason: $$\lim_{x\to1}\frac{3}{2}\Big(x-\frac{1}{x}\Big)^{2/3}=\frac{3}{2}\Big(1-\frac{1}{1}\Big)^{2/3}=\frac{3}{2}\Big(0\Big)^{2/3}=0$$ $$\lim_{x\to-1}\frac{3}{2}\Big(x-\frac{1}{x}\Big)^{2/3}=\frac{3}{2}\Big(-1-\frac{1}{-1}\Big)^{2/3}=\frac{3}{2}\Big(-1+1\Big)^{2/3}=\frac{3}{2}\Big(0\Big)^{2/3}=0$$