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

$$\lim_{x\to\infty}\frac{2x^{5/3}-x^{1/3}+7}{x^{8/5}+3x+\sqrt x}=\infty$$
We would treat these exercises just like we treat the limits of rational functions: we would divide both numerator and denominator by the highest power of $x$ in the denominator. Also remember that $$\lim_{x\to\infty}\frac{a}{x^n}=\lim_{x\to-\infty}\frac{a}{x^n}=a\times0=0\hspace{1cm}(a\in R)$$ $$A=\lim_{x\to\infty}\frac{2x^{5/3}-x^{1/3}+7}{x^{8/5}+3x+\sqrt x}=\lim_{x\to\infty}\frac{2x^{5/3}-x^{1/3}+7}{x^{8/5}+3x+x^{1/2}}$$ The highest power of $x$ in the denominator here is $x^{8/5}$, so we divide both numerator and denominator by $x^{8/5}$: $$A=\lim_{x\to\infty}\frac{2x^{5/3-8/5}-x^{1/3-8/5}+\frac{7}{x^{8/5}}}{1+3x^{1-8/5}+x^{1/2-8/5}}=\lim_{x\to\infty}\frac{2x^{1/15}-x^{-19/15}+\frac{7}{x^{8/5}}}{1+3x^{-3/5}+x^{-11/10}}$$ $$A=\lim_{x\to\infty}\frac{2x^{1/15}-\frac{1}{x^{19/15}}+\frac{7}{x^{8/5}}}{1+\frac{3}{x^{3/5}}+\frac{1}{x^{11/10}}}$$ $$A=\frac{\lim_{x\to\infty}(2x^{1/15})-0+0}{1+0+0}=\frac{\lim_{x\to\infty}(2x^{1/15})}{1}=\lim_{x\to\infty}(2x^{1/15})$$ As $x\to\infty$, $(2x^{1/15})$ will approach $\infty$ as well. So, $\lim_{x\to\infty}(2x^{1/15})=\infty$. In other words, $$A=\infty$$