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

The graph is shown below. (a) As $x\to0^+$, the graph approaches $0$. (b) As $x\to\pm\infty$, the graph approaches $3/2$. (c) As $x\to1$, the graph approaches $\infty$ from both the right and the left and as $x\to-1$, the graph approaches continuously a value of about $0.9449$.
$$y=\frac{3}{2}\Big(\frac{x}{x-1}\Big)^{2/3}$$ The graph is shown below. (a) As $x\to0^+$, the graph approaches $0$. Reason: $$\lim_{x\to0^+}\frac{3}{2}\Big(\frac{x}{x-1}\Big)^{2/3}=\frac{3}{2}\Big(\frac{0}{0-1}\Big)^{2/3}=\frac{3}{2}\times0^{2/3}=0$$ (b) As $x\to\pm\infty$, the graph approaches $3/2$. Reason: $$A=\lim_{x\to\pm\infty}\frac{3}{2}\Big(\frac{x}{x-1}\Big)^{2/3}$$ Divide the numerator and denominator both by $x$, we have $$A=\lim_{x\to\pm\infty}\frac{3}{2}\Big(\frac{1}{1-\frac{1}{x}}\Big)^{2/3}$$ $$A=\frac{3}{2}\Big(\frac{1}{1-0}\Big)^{2/3}=\frac{3}{2}\times1^{2/3}=\frac{3}{2}$$ (c) As $x\to1$, the graph approaches $\infty$ from both the right and the left and as $x\to-1$, the graph approaches continuously a value of about $0.9449$. Reason: As $x\to1$, $$x-1\to0$$ $$\frac{x}{x-1}\to\infty$$ $$\frac{3}{2}\Big(\frac{x}{x-1}\Big)^{2/3}\to\infty$$ $$\lim_{x\to-1}\frac{3}{2}\Big(\frac{x}{x-1}\Big)^{2/3}=\frac{3}{2}\Big(\frac{-1}{-1-1}\Big)^{2/3}=\frac{3}{2}\Big(\frac{1}{2}\Big)^{2/3}\approx0.9449$$