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?

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- Feb 5th 2010, 08:47 AMrodjav305Find 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 AMSoroban
Hello, rodjav305!

Quote:

Find the horizontal and vertical asymptotes of the curve: .$\displaystyle y \:=\:\frac{5e^x}{e^x-5}$

Round answers to two decimal places.

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

We have: .$\displaystyle y \:=\:\frac{5e^x}{e^x-5}$

Divide top and bottom by $\displaystyle 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: .$\displaystyle \lim_{x\to\infty}y \;=\;\lim_{x\to\infty}\left[\frac{5}{1-\frac{5}{e^x}}\right] \;=\;\frac{5}{1-0} \;=\;5$

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

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

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

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

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

- Feb 5th 2010, 11:49 AMHenryt999Dear 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 PMmr fantastic
- Feb 5th 2010, 12:56 PMRaoh
take the following function $\displaystyle f:x\mapsto \frac{1}{x}$ it's domain is $\displaystyle \mathbb{R}\setminus \left \{ 0 \right \}$.and that's means $\displaystyle x=0$ is a vertical asymptote.

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

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