# Limits involving the constant 'e'.

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?

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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.

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.

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...

Hootenanny
Staff Emeritus
Gold Member
As has been said previously,

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

However,

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

Hence the horizontal asymptote.

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