# Abs Value of X-Continuous Debate

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 defintion.

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

Last edited:

lurflurf
Homework Helper
|x| is continous 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

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
Homework Helper
Gold Member
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.

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
Homework Helper