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Math Help - Is this identically zero?

  1. #1
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    Is this identically zero?

    Is

    x\ln x

    identically zero at x=0, or does it have a singularity?

    EDIT: silly me. I didn't mean to write 'z' there, just been doing too much complex analysis recently and got into the habit of writing z everywhere.

    Sorry for the confusion!
    Last edited by scorpion007; May 27th 2010 at 11:38 PM.
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  2. #2
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by scorpion007 View Post
    Is

    x\ln x

    identically zero at z=0, or does it have a singularity?
     \lim_{x \to 0^+ } xlnx

    \lim_{x \to 0^+ } \frac{lnx}{ \frac{1}{x}}

     \lim_{x \to 0^+ } \frac{ \frac{1}{x} }{ - \frac{1}{x^2} }

     - \lim_{x \to 0^+ } \frac{ \frac{1}{x} }{ \frac{1}{x^2} }

     - \lim_{x \to 0^+ } \frac{x^2}{x}

     = 0^+

    My limits are a little bit rusty (doh!) but I think the above is good.
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  3. #3
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    The domain of the log function does not include 0, so \ln 0 is undefined, therefore 0 \cdot \ln 0 is undefined.
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  4. #4
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    Thanks guys, but you seem to be in disagreement, so I don't know which answer is correct
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  5. #5
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    Quote Originally Posted by scorpion007 View Post
    Thanks guys, but you seem to be in disagreement, so I don't know which answer is correct
    Well, we're both right. It is true that the limit as x \to 0^+ equals zero, but that has no bearing on the question you asked. I'm assuming he just misread your post.
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  6. #6
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    Ah, ok. And I guess even if a limit did exist at x=0, it still doesn't mean it won't have a singularity there, right?

    So the answer is then: It has a singularity at x=0, and is not identically 0?
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by AllanCuz View Post
     \lim_{x \to 0^+ } xlnx

    \lim_{x \to 0^+ } \frac{lnx}{ \frac{1}{x}}

     \lim_{x \to 0^+ } \frac{ \frac{1}{x} }{ - \frac{1}{x^2} }

     - \lim_{x \to 0^+ } \frac{ \frac{1}{x} }{ \frac{1}{x^2} }

     - \lim_{x \to 0^+ } \frac{x^2}{x}

     = 0^+

    My limits are a little bit rusty (doh!) but I think the above is good.
    Quote Originally Posted by drumist View Post
    Well, we're both right. It is true that the limit as x \to 0^+ equals zero, but that has no bearing on the question you asked. I'm assuming he just misread your post.
    I believe you both misinterpreted the question. It seems as though the OP is asking whether f:\mathbb{C}\to\mathbb{C}:z\mapsto z\ln(z) has a removable singularity there. The OP, in that case, has lots of explaining to do. What branch, for starters?
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  8. #8
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    No no, no complex analysis here, just purely reals. I'm not asking about removable singularities or poles or anything like that.
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  9. #9
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    By the way, asking if a function is "identically 0 for x= 0" is meaningless. A function is "identically 0" if it is 0 for all x. No, that function is not 0 for x= 0, it is undefined there.
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  10. #10
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    Quote Originally Posted by HallsofIvy View Post
    By the way, asking if a function is "identically 0 for x= 0" is meaningless. A function is "identically 0" if it is 0 for all x. No, that function is not 0 for x= 0, it is undefined there.
    I realize this is a bit off topic, but just so I have the term right, it's still meaningful to say that a function is "identically (something)" for all x in an interval (for example, f(x)=cos(x)sec(x) is identically 1 for all x\in\left(\frac{-\pi}{2},\frac{\pi}{2}\right) ), right?
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  11. #11
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    Personally, I think the adjective "identically" causes more confusion than clarity. It's clear what the statement means both with or without using the word "identically", so you might as well omit it, IMO.
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  12. #12
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    Quote Originally Posted by scorpion007 View Post
    Is

    x\ln x

    identically zero at x=0, or does it have a singularity?

    [snip]
    The function is not defined at x = 0. There is a singularity at x = 0. However, \lim_{x \to 0} (x \ln x) exists and is equal to zero (as previously shown). The point x = 0 is therefore a removable singularity (often called a 'hole').

    The function can therefore be 'fixed' if it is defined as f(x) = \left\{ \begin{array}{lr}x \ln x, & x \neq 0 \\ 0 & x = 0 \end{array} \right.

    in which case you can now say that f(0) = 0.
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