# Find the horizontal and vertical asymptotes of the curve.

• Feb 5th 2010, 08:47 AM
rodjav305
Find the horizontal and vertical asymptotes of the curve.
Find the horizontal and vertical asymptotes of the curve. (Round your answers to two decimal places.) http://www.webassign.net/cgi-bin/sym...5Ex%20-%205%29

I already solved for the y's
y=0
y=5

But I'm stuck on x?
• Feb 5th 2010, 09:30 AM
Soroban
Hello, rodjav305!

Quote:

Find the horizontal and vertical asymptotes of the curve: . $y \:=\:\frac{5e^x}{e^x-5}$
Round answers to two decimal places.

For horizontal asymptotes, find the limit of $y$ as $x\to\infty$.

We have: . $y \:=\:\frac{5e^x}{e^x-5}$
Divide top and bottom by $e^x\!:\;\;\frac{\dfrac{5e^x}{e^x}}{\dfrac{e^x}{e^x } - \dfrac{5}{e^x}} \;=\;\frac{5}{1 - \dfrac{5}{e^x}}$

Then: . $\lim_{x\to\infty}y \;=\;\lim_{x\to\infty}\left[\frac{5}{1-\frac{5}{e^x}}\right] \;=\;\frac{5}{1-0} \;=\;5$

The horizontal asymptote is: . $\boxed{y \:=\:5.00}$

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For vertical asymptotes, we want: . $\text{denominator} \,=\,0$

So we have: . $e^x-5 \:=\:0 \quad\Rightarrow\quad e^x \:=\:5 \quad\Rightarrow\quad x \:=\:\ln 5$

The vertical asymptote is: . $\boxed{x \:=\:1.61}$

• Feb 5th 2010, 11:49 AM
Henryt999
Dear Soroban
I am new to calculus and thought that was a usefull concept to "for vertical asymptotes" set the denominator to zero..
My question is how come it works out? Can I prove that somehow? Is it allways valid?
• Feb 5th 2010, 12:13 PM
mr fantastic
Quote:

Originally Posted by Henryt999
I am new to calculus and thought that was a usefull concept to "for vertical asymptotes" set the denominator to zero..
My question is how come it works out? Can I prove that somehow? Is it allways valid?

Go and look up the definition of a vertical asymptote ....
• Feb 5th 2010, 12:56 PM
Raoh
Quote:

Originally Posted by Henryt999
I am new to calculus and thought that was a usefull concept to "for vertical asymptotes" set the denominator to zero..
My question is how come it works out? Can I prove that somehow? Is it allways valid?

take the following function $f:x\mapsto \frac{1}{x}$ it's domain is $\mathbb{R}\setminus \left \{ 0 \right \}$.and that's means $x=0$ is a vertical asymptote.

• You can see that at $x=0$, $f$ is discontinuous.
• When we get closer to 0 from left and right,the curve get closer to the line $x=0$ and while doing that f takes values that go to infinity.

1. $\lim_{x\to 0^{+}}\left (\frac{1}{x} \right )=+\infty$
2. $\lim_{x\to 0^{-}}\left (\frac{1}{x} \right )=-\infty$