Chapter 2.5, Problem 42E

To determine

To find: the horizontal and vertical asymptotes of the given curve. And check the curve by using the graph.

Answer to Problem 42E

  $y=0,2\text{ and }x=\ln 5$

Explanation of Solution

Given:

The givencurve is $y=\frac{2e^x}{e^x-5}$ .

Calculation:

The given function is expressed as $y=\frac{2e^x}{e^x-5}$ .

  $\underset{x\to -\infty }{\lim }\frac{2e^x}{e^x-5}=\frac{0}{0-5}=0$

Now,going to the left, it approaches horizontal asymptotes $y=0$ .

  $\underset{x\to \infty }{\lim }\frac{2e^x}{e^x-5}\cdot \frac{e^{-x}}{e^{-x}}=\underset{x\to \infty }{\lim }\frac{2}{1-5e^{-x}}=\frac{2}{1-0}=2$

And going to the right, it approaches horizontal asymptotes $y=2$ .

Now find where the horizontal asymptote is 0.

  $\begin{array}{l}{e}^x-5\\ e^x=5\\ x=\ln 5\\ x=1.609\end{array}$

This is the vertical asymptote

Hence, the horizontal and vertical asymptotes of the given curve is $y=0,2\text{ and }x=\ln 5$ .