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Limits involving the constant 'e'.

  1. Apr 15, 2008 #1
    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?

    Thanks In Advance.
     
  2. jcsd
  3. Apr 15, 2008 #2
    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.
     
  4. Apr 15, 2008 #3

    exk

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    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.
     
  5. Apr 15, 2008 #4
    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...

    Help please!
     
  6. Apr 15, 2008 #5

    Hootenanny

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    Staff Emeritus
    Science Advisor
    Gold Member

    As has been said previously,

    [tex]\lim_{x\to\infty}2e^{x} = \infty[/tex]

    However,

    [tex]\lim_{x\to-\infty}2e^{x} = 0[/tex]

    Hence the horizontal asymptote.
     
    Last edited: Apr 15, 2008
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