Chapter 2.2, Problem 57E

To determine

To find:The reason for $\frac{g_1\left(x\right)}{g_2\left(x\right)}$ being a right end behavior model for $\frac{f_1\left(x\right)}{f_2\left(x\right)}$ .

Answer to Problem 57E

The function $\frac{g_1\left(x\right)}{g_2\left(x\right)}$ is a right end behavior model for the function $\frac{f_1\left(x\right)}{f_2\left(x\right)}$ as $\underset{x\to \infty }{\lim }\left(\frac{\frac{f_1\left(x\right)}{f_2\left(x\right)}}{\frac{g_1\left(x\right)}{g_2\left(x\right)}}\right)=1$ .

Explanation of Solution

Given information:

Suppose that $g_1\left(x\right)$ is a right end behavior model for $f_1\left(x\right)$ and $g_2\left(x\right)$ is a right end behavior model for $f_2\left(x\right)$ .

Calculation:

As given, the function $g_1\left(x\right)$ is a right end behavior model for $f_1\left(x\right)$ , then by definition $\underset{x\to \infty }{\lim }\left(\frac{f_1\left(x\right)}{g_1\left(x\right)}\right)=1$ .

It is also given that the function $g_2\left(x\right)$ is a right end behavior model for $f_2\left(x\right)$ , then by definition $\underset{x\to \infty }{\lim }\left(\frac{f_2\left(x\right)}{g_2\left(x\right)}\right)=1$ .

Now, simplify the limit of quotient.

  $\begin{array}{c}\underset{x\to \infty }{\lim }\left(\frac{\frac{f_1\left(x\right)}{f_2\left(x\right)}}{\frac{g_1\left(x\right)}{g_2\left(x\right)}}\right)=\underset{x\to \infty }{\lim }\left(\frac{f_1\left(x\right)}{g_1\left(x\right)}\times \frac{g_2\left(x\right)}{f_2\left(x\right)}\right)\\ =\underset{x\to \infty }{\lim }\left(\frac{f_1\left(x\right)}{g_1\left(x\right)}\right)\times \underset{x\to \infty }{\lim }\left(\frac{g_2\left(x\right)}{f_2\left(x\right)}\right)\\ =\underset{x\to \infty }{\lim }\left(\frac{f_1\left(x\right)}{g_1\left(x\right)}\right)\times \underset{x\to \infty }{\lim }\left(\frac{1}{\frac{f_2\left(x\right)}{g_2\left(x\right)}}\right)\\ =\underset{x\to \infty }{\lim }\left(\frac{f_1\left(x\right)}{g_1\left(x\right)}\right)\times \left(\frac{1}{\underset{x\to \infty }{\lim }\frac{f_2\left(x\right)}{g_2\left(x\right)}}\right)\end{array}$

Substitute $1$ for both $\underset{x\to \infty }{\lim }\left(\frac{f_1\left(x\right)}{g_1\left(x\right)}\right)$ and $\underset{x\to \infty }{\lim }\left(\frac{f_2\left(x\right)}{g_2\left(x\right)}\right)$ .

  $\begin{array}{c}\underset{x\to \infty }{\lim }\left(\frac{\frac{f_1\left(x\right)}{f_2\left(x\right)}}{\frac{g_1\left(x\right)}{g_2\left(x\right)}}\right)=\underset{x\to \infty }{\lim }\left(\frac{f_1\left(x\right)}{g_1\left(x\right)}\right)\times \left(\frac{1}{\underset{x\to \infty }{\lim }\frac{f_2\left(x\right)}{g_2\left(x\right)}}\right)\\ =1\times \frac{1}{1}\\ =1\end{array}$

So, the value of $\underset{x\to \infty }{\lim }\left(\frac{\frac{f_1\left(x\right)}{f_2\left(x\right)}}{\frac{g_1\left(x\right)}{g_2\left(x\right)}}\right)=1$ .

Therefore, the function $\frac{g_1\left(x\right)}{g_2\left(x\right)}$ is a right end behavior model for the function $\frac{f_1\left(x\right)}{f_2\left(x\right)}$ .