Chapter 8.2, Problem 68E

To determine

The an example of two differentiable functions $f$ and $g$ with $\underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }g\left(x\right)=\infty$ .

(a) $\underset{x\to 3}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=7$

(b) $\underset{x\to 3}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=0$

(c) $\underset{x\to 3}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\infty$

Answer to Problem 68E

All the examples are explained in the explanation.

Explanation of Solution

Given Information:

The fractions is defined as,

(a) $\underset{x\to 3}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=7$

(b) $\underset{x\to 3}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=0$

(c) $\underset{x\to 3}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\infty$

Consider the given information,

(a)

As the limit should be infinite.

So the functions are defined as,

  $f\left(x\right)=7\left(x-3\right)$ and $g\left(x\right)=x-3$

And,

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

(b)

As the limit should be infinite.

So the functions are defined as,

  $f\left(x\right)={\left(x-3\right)}^2$ and $g\left(x\right)=x-3$

Use the given limit and find the solution.

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

(c)

As the limit should be infinite.

So the functions are defined as,

  $f\left(x\right)=x-3$ and $g\left(x\right)={\left(x-3\right)}^3$

Use the given limit and find the solution.

  $\begin{array}{c}\underset{x\to 3}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to 3}{\lim }\frac{\left(x-3\right)}{{\left(x-3\right)}^3}\\ =\underset{x\to 3}{\lim }\frac{1}{{\left(x-3\right)}^2}\\ =\frac{1}{0^2}\\ =\infty \end{array}$

Hence, the required examples are explained above.