Chapter 12.4, Problem 64E

To determine

Find the asymptotes does the function appear to have?

Answer to Problem 64E

  $\boxed{y=1}$

Explanation of Solution

Given information:

Use a graphing utility to graph the function

  $f\left(x\right)=\frac{x}{\sqrt{x^2+1}}$

How many horizontal asymptotes does the function appear to have? What are the horizontal asymptotes?

Calculation:

Consider the following funnction

  $f\left(x\right)=\frac{x}{\sqrt{x^2+1}}$

To draw the graph of the funcion $f\left(x\right)=\frac{x}{\sqrt{x^2+1}}$ , use $TI-83$ plus graphing calculator. Follow the step by step procedure:

Press $\boxed{Y=}$ ,write the function and press $\boxed{ENTER}$ .

  

Press $\boxed{WINDOW}$ and adjust the $x$ and $y$ ranges. Also adjust the step sizes.

  

To draw the graph, press the $\boxed{GRAPH}$ .

  

To find asymptotes of the function $f\left(x\right)$ , calculate the limit $\underset{x\to \pm \infty }{\lim }f\left(x\right)$ .

  $\underset{x\to -\infty }{\lim }f\left(x\right)=\underset{x\to -\infty }{\lim }\frac{x}{\sqrt{x^2+1}}$

  $=\underset{x\to -\infty }{\lim }\frac{1}{\sqrt{1+\frac{1}{x^2}}}$

  $\begin{array}{l}=\frac{1}{\sqrt{1+0}}\\ =1\end{array}$

  $\underset{x\to \infty }{\lim }f(x)=\underset{x\to \infty }{\lim }\frac{x}{\sqrt{x^2+1}}$

  $\begin{array}{l}=\underset{x\to \infty }{\lim }\frac{1}{\sqrt{1+\frac{1}{x^2}}}\\ =\frac{1}{\sqrt{1+0}}\\ =1\end{array}$

Hence, the function has only one asymptote, which is $\boxed{y=1}$ .