Horizontal asymptote for e function

**klik11** · December 19th 2012, 06:01 AM

Hello,
The function is
$y = \frac{e^x + e^{-x}}{e^x - e^{-x}}$

I managed to get to the first horizontal asymptote by using lhospital's rule after I got to this:
$y = \frac{e^{2x} + 1}{e^{2x} - 1} = \frac{(e^{2x} + 1)'}{(e^{2x} - 1)'} = \frac{e^{2x} * 2}{e^{2x} * 2} = \frac{1}{1} = 1$

But then the sign of infinity which I insert doesn't matter because y will always be 1, and I know that there's another asymptote in y=-1.

How do I get to the other asymptote? Have I done anything wrong?

**MarkFL** · December 19th 2012, 07:05 AM

I would rewrite the function as:

a) For the asymptote as $x\to+\infty$

$y=\frac{1+e^{-2x}}{1-e^{-2x}}$

$\lim_{x\to+\infty}y=\frac{1+0}{1-0}=1$

b) For the asymptote as $x\to-\infty$

$y=\frac{e^{2x}+1}{e^{2x}-1}$

$\lim_{x\to-\infty}y=\frac{0+1}{0-1}=-1$

**klik11** · December 21st 2012, 01:19 PM

How did you get to that function?

**fkf** · December 21st 2012, 01:51 PM

If you're asking about a) then MarkFL2 is dividing both the numerator and denominator with $e^x$ . Doing this wont change the value of the fraction and hence is a good way to get a fraction were we easy can calculate the limit.

**MarkFL** · December 21st 2012, 03:22 PM

Yes, for:

a) $\frac{e^x+e^{-x}}{e^x-e^{-x}}\cdot\frac{e^{-x}}{e^{-x}}=\frac{1+e^{-2x}}{1-e^{-2x}}$

b) $\frac{e^x+e^{-x}}{e^x-e^{-x}}\cdot\frac{e^{x}}{e^{x}}=\frac{e^{2x}+1}{e^{2x}-1}$

**klik11** · December 22nd 2012, 02:36 AM

So I always have to get to a certain equation of the same function in order to get its limits?
How do I know when to keep trying to look for a way I can get the limit and when to let it go because it has no horizontal asymptote?

**MarkFL** · December 22nd 2012, 02:56 AM

Sometimes it is helpful to rewrite a function to more easily determine its limit, particularly limits at plus/minus infinity. When there is no horizontal asymptote, you should be able to show that the limit does not exist.

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