Chapter 11.3, Problem 3QR

Solution

(a)

To find: The vertical asymptotes for the given function $f\left(x\right)=\frac{2x^2+3}{x^2-4}$ .

The vertical asymptotes are $x=-2$ and $x=2$ .

Given information:

The given function is $f\left(x\right)=\frac{2x^2+3}{x^2-4}$ .

Formula used:

Vertical asymptote: The line $x=L$ is said to be a vertical asymptote of the given function $f\left(x\right)$ if $\underset{x\to L}{\lim }f\left(x\right)$ does not exists.

Calculation:

Consider the function $f\left(x\right)=\frac{2x^2+3}{x^2-4}$

The given function is not defined when

  $\begin{array}{c}{x}^2-4=0\\ x=\pm 2\end{array}$

So,

  $\begin{array}{c}\underset{x\to -2}{\lim }f\left(x\right)=\underset{x\to -2}{\lim }\frac{2x^2+3}{x^2-4}\\ \underset{x\to -2}{\lim }f\left(x\right)=\frac{2{\left(-2\right)}^2+3}{{\left(-2\right)}^2-4}\\ \underset{x\to -2}{\lim }f\left(x\right)=\frac{11}{0}\\ \underset{x\to -2}{\lim }f\left(x\right)=\infty \end{array}$

Clearly, $\underset{x\to -2}{\lim }f\left(x\right)$ does not exists thus $x=-2$ is a vertical asymptote of $f\left(x\right)$ .

  $\begin{array}{c}\underset{x\to 2}{\lim }f\left(x\right)=\underset{x\to 2}{\lim }\frac{2x^2+3}{x^2-4}\\ \underset{x\to 2}{\lim }f\left(x\right)=\frac{2{\left(2\right)}^2+3}{{\left(2\right)}^2-4}\\ \underset{x\to 2}{\lim }f\left(x\right)=\frac{11}{0}\\ \underset{x\to 2}{\lim }f\left(x\right)=\infty \end{array}$

Similarly, $\underset{x\to 2}{\lim }f\left(x\right)$ does not exists thus $x=2$ is also a vertical asymptote of $f\left(x\right)$ .

Therefore, the vertical asymptotes are $x=-2$ and $x=2$ .

(b)

To find: The vertical asymptotes for the given function $f\left(x\right)=\frac{2x^2+3}{x^2-4}$ .

The horizontal asymptote is $y=2$ .

Given information:

The given function is $f\left(x\right)=\frac{2x^2+3}{x^2-4}$ .

Formula used:

Horizontal asymptote: The line $y=m$ is said to be a horizontal asymptote of the function $f\left(x\right)$ if $\underset{x\to \infty }{\lim }f\left(x\right)$ or $\underset{x\to -\infty }{\lim }f\left(x\right)$ is finite.

Calculation:

  $\begin{array}{c}\underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }\frac{2x^2+3}{x^2-4}\\ \underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }\frac{x^2\left(2+\frac{3}{x^2}\right)}{x^2\left(1-\frac{4}{x^2}\right)}\\ \underset{x\to \infty }{\lim }f\left(x\right)=2\end{array}$

Thus, $y=2$ is a horizontal asymptote of $f\left(x\right)$ .