# Is this identically zero?

#### scorpion007

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:

#### AllanCuz

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.

• scorpion007

#### drumist

The domain of the log function does not include 0, so $$\displaystyle \ln 0$$ is undefined, therefore $$\displaystyle 0 \cdot \ln 0$$ is undefined.

• scorpion007

#### scorpion007

Thanks guys, but you seem to be in disagreement, so I don't know which answer is correct #### drumist

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.

#### scorpion007

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

MHF Hall of Honor
$$\displaystyle \lim_{x \to 0^+ } xlnx$$

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

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

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

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

$$\displaystyle = 0^+$$

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?

#### scorpion007

No no, no complex analysis here, just purely reals. I'm not asking about removable singularities or poles or anything like that.

#### HallsofIvy

MHF Helper
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.

• scorpion007

#### Diamondlance

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?