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Math Help - Limits of e functions

  1. #1
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    Limits of e functions

    Help!

    How do I find the limits for the following:

    lim x-> 0+ of e^(1/x)

    Lim x->-inf of (1-e^x)/(1+e^x).

    I can't figure these out...please help!

    Jeannine
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  2. #2
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    Hello, Jeannine!

    Here's the second one . . .


    \lim_{x\to\text{-}\infty}\,\frac{1-e^x}{1 + e^x}
    Divide top and bottom by e^x

    \lim_{x\to\text{-}\infty}\frac{\dfrac{1}{e^x} - \dfrac{e^x}{e^x}}{\dfrac{1}{e^x} + \dfrac{e^x}{e^x}} \;=\;\lim_{x\to\text{-}\infty}\frac{\dfrac{1}{e^x} - 1}{\dfrac{1}{e^x} + 1} \;=\;\frac{0-1}{0+1} \;=\;-1

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  3. #3
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    The answer in the back of my book is not -1...

    It says +1.

    Jeannine
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  4. #4
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    Your text is correct.
    You should know that \color{red}\lim _{x \to  - \infty } e^x  = 0

    Therefore you can see that \frac {1-0} {1+0} = 1
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  5. #5
    Super Member Showcase_22's Avatar
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    what? So was Soroban wrong? I couldn't see anything wrong with that method!

    \lim_{x\rightarrow 0}e^{1/x}
    Does your book say from which side x is approaching from? If x is positive, it's +\infty and if x is negative it's -\infty.

    Someone good at analysis might find a way around this though.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    what? So was Soroban wrong? I couldn't see anything wrong with that method!



    Does your book say from which side x is approaching from? If x is positive, it's +\infty and if x is negative it's -\infty.

    Someone good at analysis might find a way around this though.
    \lim_{x \to - \infty} \frac 1{e^x} = \lim_{x \to - \infty} e^{-x} = \infty not zero. i think Soroban forgot about the negative infinity.

    one-sided limits don't exist if the limit is infinite. how could you approach infinity from the right? or - infinity from the left?
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  7. #7
    Super Member Showcase_22's Avatar
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    one-sided limits don't exist if the limit is infinite. how could you approach infinity from the right? or - infinity from the left?
    I meant x was approaching zero from the right or left. I should have explained it better.

    \lim_{x \rightarrow 0+} \frac {1}{x}=+ \infty


    \lim_{x \rightarrow 0-} \frac {1}{x}=- \infty

    (I think I wrote that out correctly :s)

    Then this would make:

    e^{(+ \infty)}= \infty

    e^{(- \infty)}=0

    Oops! Wrong in my previous post!

    Am I making the problem too difficult?

    \lim_{x \rightarrow -\infty} e^{(-x)}
    I didn't notice it was - \infty. I thought it was + \infty. Oops!
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    I didn't notice it was - \infty. I thought it was + \infty. Oops!
    yes. i suppose Soroban forgot that fact as well, as i said in my last post
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  9. #9
    Senior Member bkarpuz's Avatar
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    You may also do substitution in limit as follows:
    \lim\limits_{x\to-\infty}\mathrm{e}^{x}\stackrel{y\mapsto-x}{=}\lim\limits_{y\to\infty}\mathrm{e}^{-y}=\lim\limits_{y\to\infty}\frac{1}{\mathrm{e}^{y}  }=0.
    Last edited by bkarpuz; September 29th 2008 at 11:24 AM.
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  10. #10
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    Look all of you; the discussion was about the second of the two problems not the fiist: \lim _{x \to  - \infty } \frac{{1 - e^x }}{{1 + e^x }}.
    That limit can easily be done by simple inspection knowing that \lim _{x \to  - \infty } e^x  = 0.

    The first problem was \lim _{x \to 0^ +  } e^{\frac{1}{x}} which is also easy to see that the limit does exist or is +\infty depending on the textbook.
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