# Limits involving the constant 'e'.

1. Apr 15, 2008

### Morphayne

Hi. I'm confused about calculating the limit of a function involving Euler's number. I need to know the proper way to find the limit so I can determine the equation of the horizontal asymptote of the function. I do know, in my head what the graph looks like so I know that there is a horizontal asymptote is y=0. I just have to show my work, thats where I'm stuck.

Here is the problem:

lim(x->infinity)(2e^x)

My attempt:

lim(x->infinity)(2e^x) = lim(x->infinity)(2/e^-x)

=lim(x->infinity)(2/e^-infinity)

=lim(x->infinity)(2/0)

=0 Therefore: y=0

I'm not sure if I'm doing the problem the right way. If I did somehow get it right, can someone please give me a brief description on the proper method?

2. Apr 15, 2008

### steelphantom

Isn't the lim(2*e^x) = infinity? As x approaches infinity, so does e^x. You don't even really need those last couple steps, because lim(2e^x) = 2e^(infinity) = infinity.

3. Apr 15, 2008

### exk

Morphayne,

2/0 doesn't equal 0, it's "equal" to infinity, but in reality it's undefined.

Your original limit is equal to infinity for the reason that steelphantom mentioned.

4. Apr 15, 2008

### Morphayne

Well, the answer in the textbook is says that there is a horizontal asymptote at y=0 which means that this limit somehow equals zero...

5. Apr 15, 2008

### Hootenanny

Staff Emeritus
As has been said previously,

$$\lim_{x\to\infty}2e^{x} = \infty$$

However,

$$\lim_{x\to-\infty}2e^{x} = 0$$

Hence the horizontal asymptote.

Last edited: Apr 15, 2008