Abs Value of X-Continuous Debate

[Tom McCurdy]

Question:
Is $$f(x)=\mid{x}\mid$$ continuous?

I have been looking online and got a few different answers. My calc B.C. teacher last year claimed that $$f(x)=\mid{x}\mid$$ is continuous everywhere except at x=0. My current 115 teacher maintains that anyone under that impression deserves to be boiled in their own pudding. It seems to me that it would make sense that it is continuous, but that is from a conceptual view rather than mathmatical definition.

If anyone has any proof please tell me and provide link to site confirming it. Especially if it is not continuous at 0.

 

[lurflurf]

|x| is continuous everywhere (including 0)
and has a continuous derivatives of all orders everywhere except 0.
The proof is obvious since
|x|=-x x<0
|x|=x x>0
x and -x clearly have continuous derivatives of all orders
|0+|=|0-|=|0|=0 so we have |x| is continuous everywhere
f'(x)=-1 x<0
f'(x)=1 x>0
so f'(x) does not exist at 0 so is not continuous
likewise higher derivatives

 

[master_coda]

Alright, let $f(x)=\lvert x\rvert$. So, given any $\varepsilon>0$ take $\delta=\varepsilon$. Then if $\lvert x-0\vert<\delta$ then $\lvert f(x)-f(0)\rvert=\lvert f(x)\rvert<\varepsilon$ and so $\lim_{x\rightarrow 0}f(x)=0=f(0)$ and therefore f(x) is continuous at x=0.

 

[James R]

Is it true that a function f(x) is continuous at x=a if:

$$\lim_{x \rightarrow a+} f(x) = \lim_{x \rightarrow a-} f(x)$$?

Clearly, this holds for f(x) = |x| at x=0.

 

[Crosson]

Is it true that a function f(x) is continuous at x=a if:

The other condition is obvious but necessary, we require the limit from both sides to exist and be equal to the value of the function at that point.

 

[Galileo]

Your former teacher may have been confused with differentiability, meaning he probably had a bad day.
Or he's just plain dumb, I dunno.

 

[HallsofIvy]

Or maybe you misunderstood! f(x)= |x| is continuous at x= 0 but not differentiable there.

 

[Tom McCurdy]

Thanks, it probably is my memory since it was from last year...