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

$$\lim_{x\to\infty}\frac{x^{2/3}+x^{-1}}{x^{2/3}+\cos^2 x}=1$$
*Remember that $\lim_{x\to\pm\infty}\frac{a}{x^n}=a\lim_{x\to\pm\infty}\frac{1}{x^n}=a\times0=0$ $$A=\lim_{x\to\infty}\frac{x^{2/3}+x^{-1}}{x^{2/3}+\cos^2 x}$$ Divide both numerator and denominator by the highest degree of $x$ in the denominator, which is $x^{2/3}$ here: $$A=\lim_{x\to\infty}\frac{1+\frac{x^{-1}}{x^{2/3}}}{1+\frac{\cos^2x}{x^{2/3}}}=\lim_{x\to\infty}\frac{1+\frac{1}{x\times x^{2/3}}}{1+\frac{\cos^2x}{x^{2/3}}}=\lim_{x\to\infty}\frac{1+\frac{1}{x^{5/3}}}{1+\frac{\cos^2x}{x^{2/3}}}$$ $$A=\frac{1+0}{1+\lim_{x\to\infty}\frac{\cos^2x}{x^{2/3}}}=\frac{1}{1+\Big(\lim_{x\to\infty}\frac{\cos x}{x^{1/3}}\Big)^2}$$ We know that $-1\le\cos x\le1$, hence $\frac{-1}{x^{1/3}}\le\frac{\cos x}{x^{1/3}}\le\frac{1}{x^{1/3}}$ (because $x\to\infty$, we assume $x\gt0$ here) Furthermore, $\lim_{x\to\infty}-1/x^{1/3}=\lim_{x\to\infty}1/x^{1/3}=0$ Therefore, according to Sandwich Theorem, $\lim_{x\to\infty}(\cos x/x^{1/3})=0$ $$A=\frac{1}{1+0^2}=1$$