Is this identically zero?

Jul 2006
364
44
Is

\(\displaystyle x\ln x\)

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:
Apr 2010
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Canada
Is

\(\displaystyle x\ln x\)

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

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

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

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

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

\(\displaystyle = 0^+ \)

My limits are a little bit rusty (doh!) but I think the above is good.
 
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Jan 2010
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The domain of the log function does not include 0, so \(\displaystyle \ln 0\) is undefined, therefore \(\displaystyle 0 \cdot \ln 0\) is undefined.
 
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Jul 2006
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Thanks guys, but you seem to be in disagreement, so I don't know which answer is correct :)
 
Jan 2010
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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 \(\displaystyle x \to 0^+\) equals zero, but that has no bearing on the question you asked. I'm assuming he just misread your post.
 
Jul 2006
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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?
 

Drexel28

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\(\displaystyle \lim_{x \to 0^+ } xlnx \)

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

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

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

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

\(\displaystyle = 0^+ \)

My limits are a little bit rusty (doh!) but I think the above is good.
Well, we're both right. It is true that the limit as \(\displaystyle x \to 0^+\) 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 \(\displaystyle f:\mathbb{C}\to\mathbb{C}:z\mapsto z\ln(z)\) has a removable singularity there. The OP, in that case, has lots of explaining to do. What branch, for starters?
 
Jul 2006
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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.
 

HallsofIvy

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Apr 2005
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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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Oct 2008
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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.
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, \(\displaystyle f(x)=cos(x)sec(x)\) is identically 1 for all \(\displaystyle x\in\left(\frac{-\pi}{2},\frac{\pi}{2}\right)\) ), right?