Chapter 12.4, Problem 32E

To determine

To estimate the horizontal asymptotes and find limit at infinity for the function $y=\frac{2x+1}{x^2-1}$

using graphing utility.

Answer to Problem 32E

The horizontal asymptotes $y=0$ and limit of $y$ is $0$ as $x\to \pm \infty$

Explanation of Solution

Given information :

Given function $y=\frac{2x+1}{x^2-1}$

Graph:

  

Interpretation: from above graph, we observe that the horizontal asymptote is $y=0$ and also $y\to 0$ as $x\to \pm \infty$

To find limit of $y=\frac{2x+1}{x^2-1}$ at infinity.

  $\begin{array}{l}.i.e,\underset{x\to \pm \infty }{\lim }y=\underset{x\to \pm \infty }{\lim }\frac{2x+1}{x^2-1}\\ =\underset{x\to \pm \infty }{\lim }\pm \frac{x\left(2+\frac{1}{x}\right)}{x^2\left(-\frac{1}{x^2}+1\right)}\\ =\underset{x\to \pm \infty }{\lim }\frac{2}{x}\left(\because \underset{x\to \pm \infty }{\lim }\frac{1}{x^2}=0,\underset{x\to \pm \infty }{\lim }\frac{1}{x}=\right)\\ \therefore \underset{x\to \pm \infty }{\lim }y=0\end{array}$

Which is the value of $y$ for which the horizontal asymptotes exists.

Hence, the horizontal asymptotes $y=0$ and limit of $y$ is $0$ as $x\to \pm \infty$