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Thread: 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 . . .


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

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

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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 $\displaystyle \color{red}\lim _{x \to - \infty } e^x = 0$

    Therefore you can see that $\displaystyle \frac {1-0} {1+0} = 1$
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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!

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

    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 $\displaystyle +\infty$ and if x is negative it's $\displaystyle -\infty$.

    Someone good at analysis might find a way around this though.
    $\displaystyle \lim_{x \to - \infty} \frac 1{e^x} = \lim_{x \to - \infty} e^{-x} = \infty$ 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.

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


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

    (I think I wrote that out correctly :s)

    Then this would make:

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

    $\displaystyle e^{(- \infty)}=0$

    Oops! Wrong in my previous post!

    Am I making the problem too difficult?

    $\displaystyle \lim_{x \rightarrow -\infty} e^{(-x)}$
    I didn't notice it was $\displaystyle - \infty$. I thought it was $\displaystyle + \infty$. 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 $\displaystyle - \infty$. I thought it was $\displaystyle + \infty$. 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:
    $\displaystyle \lim\limits_{x\to-\infty}\mathrm{e}^{x}\stackrel{y\mapsto-x}{=}\lim\limits_{y\to\infty}\mathrm{e}^{-y}=\lim\limits_{y\to\infty}\frac{1}{\mathrm{e}^{y} }=0$.
    Last edited by bkarpuz; Sep 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: $\displaystyle \lim _{x \to - \infty } \frac{{1 - e^x }}{{1 + e^x }}$.
    That limit can easily be done by simple inspection knowing that $\displaystyle \lim _{x \to - \infty } e^x = 0$.

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