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Math Help - limits

  1. #1
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    limits

    Hi all,

    Just wondering if any one can show some working out and explain the following problem:

    lim x --> 0+ ( sqrt(x)*ln(x)) =

    I know the answer is 0 and i guess you can say because 0* infinity = 0. However, i don't feel that is the correct working out. Could some one please show some working out as well as explaining it.

    Thank you in advance,
    ArTiCk
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by ArTiCK View Post
    Hi all,

    Just wondering if any one can show some working out and explain the following problem:

    lim x --> 0+ ( sqrt(x)*ln(x)) =

    I know the answer is 0 and i guess you can say because 0* infinity = 0. However, i don't feel that is the correct working out. Could some one please show some working out as well as explaining it.

    Thank you in advance,
    ArTiCk
    Put y=1/x, then:

    \lim_{x \to 0+} \sqrt{x}\ln(x) =-~\lim_{y \to \infty} \frac{\ln(y)}{\sqrt{y}}


    There are a number of ways of tacking this, one (not the best necessarily but that you may be familiar with) is L'Hopitals rule.

    RonL
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ArTiCK View Post
    Hi all,

    Just wondering if any one can show some working out and explain the following problem:

    lim x --> 0+ ( sqrt(x)*ln(x)) =

    I know the answer is 0 and i guess you can say because 0* infinity = 0. However, i don't feel that is the correct working out. Could some one please show some working out as well as explaining it.

    Thank you in advance,
    ArTiCk
    No! 0*infinity makes no sense!

    now, are you in calculus. L'Hopital's rule seems pretty easy here *covers his ears to drown out the boos coming from the other mathematicians*

    note that \lim_{x \to 0^+} \sqrt{x} \ln x = \lim_{x \to 0^+} \frac {\ln x}{x^{-1/2}} = - \lim_{x \to 0^+} \frac {- \ln x}{x^{-1/2}}

    this satisfies the conditions for using L'Hoptial's.

    Apply L'Hopital's and we get - \lim_{x \to 0^+} 2 \frac {\frac 1x}{x^{-3/2}}

    simplify this and take the limit
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  4. #4
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    Quote Originally Posted by Jhevon View Post
    No! 0*infinity makes no sense!

    now, are you in calculus. L'Hopital's rule seems pretty easy here *covers his ears to drown out the boos coming from the other mathematicians*

    note that \lim_{x \to 0^+} \sqrt{x} \ln x = \lim_{x \to 0^+} \frac {\ln x}{\sqrt{x}} = - \lim_{x \to 0^+} \frac {- \ln x}{\sqrt{x}}
    Somethings gone wrong here!

    RonL
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Somethings gone wrong here!

    RonL
    I already changed it. i wanted to have \frac 1{\sqrt{x}} in the denominator, btu it looked messy and i started editing...long story
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by ArTiCK View Post
    Hi all,

    Just wondering if any one can show some working out and explain the following problem:

    lim x --> 0+ ( sqrt(x)*ln(x)) =

    I know the answer is 0 and i guess you can say because 0* infinity = 0. However, i don't feel that is the correct working out. Could some one please show some working out as well as explaining it.

    Thank you in advance,
    ArTiCk
    I have to do it...I dont know advanced you are, but if you have gotten to infinite series

    \ln(x)=\sum_{n=0}^{\infty}\frac{(-1)^n(x-1)^{n+1}}{n+1}, \forall{x}\in[0,2)
    and since 0^{+} is contained in this
    \therefore\sqrt{x}\ln(x)=\sqrt{x}\cdot\sum_{n=0}^{  \infty}\frac{(-1)^n(x-1)^{n+1}}{n+1}

    as you can see every term will have an x in it, thus direct substitution will yield 0+0+0+...+0=0

    \therefore\lim_{x\to{0^{+}}}\sqrt{x}\ln(x)=0

    EDIT: Also, are you aware of the conept of higher order infinitesimals with limits, \sqrt{x}\succ\ln(x)?
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  7. #7
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    Quote Originally Posted by Mathstud28 View Post
    I have to do it...I dont know advanced you are, but if you have gotten to infinite series

    \ln(x)=\sum_{n=0}^{\infty}\frac{(-1)^n(x-1)^{n+1}}{n+1}, \forall{x}\in[0,2)
    and since 0^{+} is contained in this
    \therefore\sqrt{x}\ln(x)=\sqrt{x}\cdot\sum_{n=0}^{  \infty}\frac{(-1)^n(x-1)^{n+1}}{n+1}

    as you can see every term will have an x in it, thus direct substitution will yield 0+0+0+...+0=0

    \therefore\lim_{x\to{0^{+}}}\sqrt{x}\ln(x)=0

    EDIT: Also, are you aware of the conept of higher order infinitesimals with limits, \sqrt{x}\succ\ln(x)?
    What you have here is:

    \lim_{x \to 0+} \sqrt{x}\ln(x)=\lim_{x \to 0+} \lim_{k \to \infty} \sum_{n=0}^k \sqrt{x}~\frac{(-1)^n(x-1)^{n+1}}{n+1}

    and to complete this you want to conclude that:

    \lim_{x \to 0+} \lim_{k \to \infty}\sum_{n=0}^k \sqrt{x}~\frac{(-1)^n(x-1)^{n+1}}{n+1}=\lim_{k \to \infty} \lim_{x \to 0+} \sum_{n=0}^k\sqrt{x}~\frac{(-1)^n(x-1)^{n+1}}{n+1}

    So to complete this you need to justify this last step.

    RonL
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    What you have here is:

    \lim_{x \to 0+} \sqrt{x}\ln(x)=\lim_{x \to 0+} \lim_{k \to \infty} \sum_{n=0}^k \sqrt{x}~\frac{(-1)^n(x-1)^{n+1}}{n+1}

    and to complete this you want to conclude that:

    \lim_{x \to 0+} \lim_{k \to \infty}\sum_{n=0}^k \sqrt{x}~\frac{(-1)^n(x-1)^{n+1}}{n+1}=\lim_{k \to \infty} \lim_{x \to 0+} \sum_{n=0}^k\sqrt{x}~\frac{(-1)^n(x-1)^{n+1}}{n+1}

    So to complete this you need to justify this last step.

    RonL
    Would I actually have to prove it or could I just say that the order of limits is irrelevant here?
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  9. #9
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    Quote Originally Posted by Mathstud28 View Post
    Would I actually have to prove it or could I just say that the order of limits is irrelevant here?
    It's essential to attend to these details.

    As you become more advanced in your understanding of series, you will appreciate the importance of this.
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by mr fantastic View Post
    It's essential to attend to these details.

    As you become more advanced in your understanding of series, you will appreciate the importance of this.
    *ahem* I do understand the importance, I simply was pointing out that at this point it did not seem pertinent,
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  11. #11
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    Quote Originally Posted by Mathstud28 View Post
    *ahem* I do understand the importance, I simply was pointing out that at this point it did not seem pertinent,
    It is extremely pertinant as it is the key thing to proving the required result this way.

    I do not know without doing a bit of reserch what would justify changing the limits in this case.

    (or rather I do know what would justify changing the order of the limits, I just think that it would not be worth the effort of showing that it applied in this case, given that we have simpler methods available)

    RonL
    Last edited by CaptainBlack; May 30th 2008 at 10:42 PM.
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