4.6 Limits at infinity and asymptotes

1. Using the algebraic limit laws, we have $\underset{x\to \infty }{\text{lim}}\left(5-\frac{2}{x^2}\right)=\underset{x\to \infty }{\text{lim}}5-2\left(\underset{x\to \infty }{\text{lim}}\frac{1}{x}\right).\left(\underset{x\to \infty }{\text{lim}}\frac{1}{x}\right)=5-2·0=5.$
   Similarly, $\underset{x\to \infty }{\text{lim}}f\left(x\right)=5.$ Therefore, $f\left(x\right)=\frac{5-2}{x^2}$ has a horizontal asymptote of $y=5$ and $f$ approaches this horizontal asymptote as $x\to \text{±}\infty$ as shown in the following graph.
   

   

2. Since $\mathrm{-1}\le \text{sin}x\le 1$ for all $x,$ we have
   
   $\frac{\mathrm{-1}}{x}\le \frac{\text{sin}x}{x}\le \frac{1}{x}$
   
   for all $x\ne 0.$ Also, since
   
   $\underset{x\to \infty }{\text{lim}}\frac{\mathrm{-1}}{x}=0=\underset{x\to \infty }{\text{lim}}\frac{1}{x},$
   
   we can apply the squeeze theorem to conclude that
   
   $\underset{x\to \infty }{\text{lim}}\frac{\text{sin}x}{x}=0.$
   
   Similarly,
   
   $\underset{x\to \text{−}\infty }{\text{lim}}\frac{\text{sin}x}{x}=0.$
   
   Thus, $f\left(x\right)=\frac{\text{sin}x}{x}$ has a horizontal asymptote of $y=0$ and $f\left(x\right)$ approaches this horizontal asymptote as $x\to \text{±}\infty$ as shown in the following graph.
   

   

3. To evaluate $\underset{x\to \infty }{\text{lim}}{\text{tan}}^{\mathrm{-1}}\left(x\right)$ and $\underset{x\to \text{−}\infty }{\text{lim}}{\text{tan}}^{\mathrm{-1}}\left(x\right),$ we first consider the graph of $y=\text{tan}\left(x\right)$ over the interval $\left(\text{−}\pi \text{/}2,\pi \text{/}2\right)$ as shown in the following graph.
   

   

Since

it follows that

Similarly, since

it follows that

As a result, $y=\frac{\pi }{2}$ and $y=-\frac{\pi }{2}$ are horizontal asymptotes of $f\left(x\right)={\text{tan}}^{\mathrm{-1}}\left(x\right)$ as shown in the following graph.