Chapter 2.6, Problem 19E

Solution

The horizontal and vertical asymptotes of $f\left(x\right)$ and describe the corresponding behaviour using limits.

The line is $y=2$ horizontal asymptotes and there is no vertical asymptote.

Given:

The function $f\left(x\right)=\frac{2x^2-1}{x^2+3}$ .

Concept Used:

If a polynomial function in the form $y=\frac{f\left(x\right)}{g\left(x\right)}=\frac{\left(a_n{x}^n+\mathrm{....}\right)}{\left(b_m{x}^m+\mathrm{....}\right)}$ , then, vertical asymptotes is given by real zeros of denominator provided it should not be zero of numerator.

And,

The end behaviour asymptotes given by $\underset{x\to {\infty }^{-}}{\lim }\frac{f\left(x\right)}{g\left(x\right)}\text{ or }\underset{x\to {\infty }^{+}}{\lim }\frac{f\left(x\right)}{g\left(x\right)}$ is horizontal asymptotes.

The condition can be concluded as,

1) If $n<m$ , the end behaviour is horizontal asymptotes that is $y=0$ ,

2) If $n=m$ the end behaviour is horizontal asymptotes is horizontal asymptotes that is $y=\frac{a_n}{b_n}$

3) If $n>m$ ,the quotient $q\left(x\right)$ of $f\left(x\right)/g\left(x\right)$ is the end behaviour asymptotes, where $q\left(x\right)$ is given by $f\left(x\right)=g\left(x\right)q\left(x\right)+r\left(x\right)$ with $r\left(x\right)$ is remainder of $f\left(x\right)/g\left(x\right)$ . Hence, no horizontal asymptotes.

Calculation:

Consider the function, $f\left(x\right)=\frac{2x^2-1}{x^2+3}$ .

The denominator is always greater than zero and there is no real roots.

Hence, no vertical asymptotes.

To find end behaviour asymptotes, compute, $\underset{x\to \infty }{\lim }f\left(x\right)$ .

So, rewrite $f\left(x\right)=\frac{2x^2-1}{x^2+3}$ using long division,

  $x^2+3\stackrel{2}{\overline{)\begin{array}{l}2x^2-1\\ -\underset{¯}{\left(2x^2+6\right)}\\ -7\end{array}}}$

So, $\left(2x^2-1\right)=2\left(x^2+3\right)-7$

Now, use $\left(2x^2-1\right)=2\left(x^2+3\right)-7$ in $f\left(x\right)=\frac{2x^2-1}{x^2+3}$ and simplify:

  $\begin{array}{c}f\left(x\right)=\frac{2\left(x^2+3\right)-7}{x^2+3}\\ =\frac{2\left(x^2+3\right)}{x^2+3}-\frac{7}{x^2+3}\\ =2-\frac{7}{x^2+3}\end{array}$

Thus, $f\left(x\right)=2-\frac{7}{x^2+3}$ .

Since, for, $x$ approaches to infinity $f\left(x\right)$ approaches to $2$ .

That is,

  $\begin{array}{c}\underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }\left(2-\frac{7}{x^2+3}\right)\\ =2\end{array}$

Hence, end behaviour is horizontal asymptotes $y=2$ .

Conclusion:

The line is $y=2$ horizontal asymptotes and there is no vertical asymptote.