## Calculus: Early Transcendentals 8th Edition

The asymptotes are: 1) Line $x=0$ is vertical asymptote from the left and $y$ goes to $+\infty$; 2) Line $x=0$ is vertical asymptote from the right and $y$ goes to $+\infty$; 3) Line $y=0$ is horizontal asymptote when $x\to-\infty$; 4) Line $y=0$ is horizontal asymptote when $x\to +\infty$.
From the graph below we suspect that the asymptotes are: 1) Line $x=0$ is vertical asymptote from the left and $y$ goes to $+\infty$; 2) Line $x=0$ is vertical asymptote from the right and $y$ goes to $+\infty$; 3) Line $y=0$ is horizontal asymptote when $x\to-\infty$; 4) Line $y=0$ is horizontal asymptote when $x\to +\infty$. Let's prove this: 1) $$\lim_{x\to0^-}\frac{\cos^2x}{x^2}=\left[\frac{\cos^2 0^-}{(0^-)^2}\right]=\left[\frac{1}{0^+}\right]=+\infty$$ which is what is needed. 2) $$\lim_{x\to 0^+}\frac{\cos^2 x}{x^2}=\left[\frac{\cos^20^+}{(0^+)^2}\right]=\left[\frac{1}{0^+}\right]=+\infty$$ which is what is needed. 3) $$\lim_{x\to -\infty}\frac{\cos^2x}{x^2}=\left[\frac{\text{something that is always between -1 and 1}}{(-\infty)^2}\right]=\left[\frac{\text{something that is always between -1 and 1}}{+\infty}\right]=0.$$ which is what is needed. 4) $$\lim_{x\to +\infty}\frac{\cos^2x}{x^2}=\left[\frac{\text{something that is always between -1 and 1}}{(+\infty)^2}\right]=\left[\frac{\text{something that is always between -1 and 1}}{+\infty}\right]=0.$$ which is what is needed. 