Chapter 11.4, Problem 66E

To determine

To calculate:

No of asymptote does given function has and show the function with the help of graph.

Answer to Problem 66E

Given function $f\left(x\right)=\frac{x}{\sqrt{x^2+1}}$ has only one asymptote that is $y=1$

Explanation of Solution

Given information:

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

Calculation:

For completing the table of the sequence,

Consider the given functions,

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

To determining the asymptote of the function, first calculate the limit $\underset{x\to \pm \infty }{\lim }f\left(x\right)$

  $\begin{array}{l}\underset{x\to \pm \infty }{\lim }f\left(x\right)=\underset{x\to \pm \infty }{\lim }\frac{x}{\sqrt{x^2+1}}\\ =\underset{x\to \pm \infty }{\lim }\frac{1}{\sqrt{1+\frac{1}{x^2}}}\\ =\frac{1}{\sqrt{1+0}}\\ =1\\ \underset{x\to \pm \infty }{\lim }f\left(x\right)=\underset{x\to \pm \infty }{\lim }\frac{x}{\sqrt{x^2+1}}\\ =\underset{x\to \pm \infty }{\lim }\frac{x}{\sqrt{1+\frac{1}{x^2}}}\\ =\frac{1}{\sqrt{1+0}}\\ =1\end{array}$

Therefore, the function has only one asymptote that is $y=1$

The graph is shown with the help of desmos graphing utility as below for the given function $f\left(x\right)=\frac{x}{\sqrt{x^2+1}}$ :