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Thread: Some easy ->infinity limits I can't solve

  1. #1
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    Some easy ->infinity limits I can't solve

    Hi,

    I'm having some problems with calculating quite a few limits, so I guess there's a concept I'm not getting...

    Fe:
    $\displaystyle \lim_{x\to\infty}x^{1/x}$
    Or:
    $\displaystyle \lim_{x\to\infty}x^4e^{-x^2}$

    $\displaystyle \lim_{x\to\infty}\frac{e^x}{\sqrt(x)}$

    How should I solve these?

    Thanks!
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  2. #2
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by superdl View Post
    Hi,

    I'm having some problems with calculating quite a few limits, so I guess there's a concept I'm not getting...

    Fe:
    $\displaystyle \lim_{x\to\infty}x^{\frac {1}{x}}$
    Or:
    $\displaystyle \lim_{x\to\infty}x^4e^{-x^2}$

    $\displaystyle \lim_{x\to\infty}\frac{e^x}{\sqrt{x}}$

    How should I solve these?

    Thanks!
    first one is one == 1
    (hint: it's $\displaystyle \infty ^0$ so u have to ... == $\displaystyle \displaystyle e ^{\lim_{x\to\infty}\frac {log(x)}{x}} $ and so on ... )

    $\displaystyle \displaystyle \lim_{x\to\infty}x^{\frac {1}{2}} = 1$

    second is zero (first use L'Hospitals rule .... )
    (hint: after first use of rule you should get $\displaystyle \displaystyle \lim_{x\to\infty} \frac {2x^2}{e^{x^2}}$ and after second use you will get $\displaystyle \displaystyle \lim_{x\to\infty} \frac {2}{e^{x^2}}$ ..... )

    $\displaystyle \displaystyle \lim_{x\to\infty}\frac{e^x}{\sqrt{x}}=0$

    third one is infinity (first use L'Hospitals rule .... )
    (hint: after use of rule you will get $\displaystyle \displaystyle \lim_{x\to\infty} 2 e^x \sqrt{x}$ .... )

    $\displaystyle \displaystyle \lim_{x\to\infty}\frac{e^x}{\sqrt{x}} = \infty $
    Last edited by yeKciM; Aug 12th 2010 at 01:48 PM.
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  3. #3
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    That's what I thought, but I also read that infinity^0 = undefined ?
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  4. #4
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    I see you've edited your post.

    The first one: could you elaborate a bit further please? :$

    I understand the second one. I was doing l'Hospitals, but I didn't notice I could rule out the 2x every time (stupid mistake)
    I also understand the third one.

    Thanks for your help so far!
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  5. #5
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by superdl View Post
    I see you've edited your post.

    The first one: could you elaborate a bit further please? :$

    I understand the second one. I was doing l'Hospitals, but I didn't notice I could rule out the 2x every time (stupid mistake)
    I also understand the third one.

    Thanks for your help so far!
    when you get form like $\displaystyle \infty^0$ u can transform it like :

    $\displaystyle \displaystyle \lim_{x\to \infty } x^{\frac{1}{x}} = e^{\displaystyle \lim_{x\to \infty }}\frac {log{(x)}}{x}}$

    now u use L'Hospital's rule again and you'll get :

    $\displaystyle \displaystyle e^{\displaystyle \lim_{x\to \infty } \frac {1}{x}} = 1 $
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  6. #6
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    I did some more researching and I get it know. Although, shouldn't it be ln instead of log ?
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  7. #7
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    Quote Originally Posted by superdl View Post
    Although, shouldn't it be ln instead of log ?
    There seems to be a trend, particularly in US mathematics, to use $\displaystyle \log(x)$ in place of $\displaystyle \ln(x)$.
    The late Leonard Gillman (Rings of Continuous Functions) and past president of the MAA was most vocal about this.
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  8. #8
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    Do you mean that log should be used as the notation for the logarithm with base e then?

    'cause $\displaystyle e^{ln a} = a$, and $\displaystyle 10^{log a} = a$, but $\displaystyle e^{log a} \not= a$

    (Ps: As you may have noticed, I'm not from the US; I live in Belgium)
    Last edited by superdl; Aug 13th 2010 at 05:27 AM.
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