How would you show that the absolute value function $\displaystyle f(x) := |x|$ is continuous at every point $\displaystyle c$ in $\displaystyle \mathbb{R}$

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- Feb 24th 2010, 07:43 PMCrazyCat87Absolute Value Function
How would you show that the absolute value function $\displaystyle f(x) := |x|$ is continuous at every point $\displaystyle c$ in $\displaystyle \mathbb{R}$

- Feb 24th 2010, 07:44 PMDrexel28
- Feb 25th 2010, 03:49 AMHallsofIvy
Well, if he is asking this, he probably doesn't know about "Lipschitz"- or "uniformly continuous"!

CrazyCat87, The**definition**of f(x) continuous at x= a is that f(a) exist, $\displaystyle \lim_{x\to a} f(x)$ exist, and $\displaystyle \lim_{x\to a} f(x)= f(a)$. Typically, showing that the two are equal requires showing they exist so that is enough.

Here, if x> 0, f(x)= |x|= x. If a> 0, we can take $\displaystyle \delta< a/2$ so that as long as $\displaystyle |x-a|<\delta$, x> 0 and $\displaystyle |f(x)- f(a)|= |x- a|< \epsilon$ will be true as long as $\displaystyle \delta$ is less than the smaller of a/2 and $\displaystyle \epsilon$.

If x< 0, f(x)= |x|= -x. If a< 0, we can take $\displaystyle \delta< -a/2$ so that as long as $\displaystyle |x-a|< \delta$, x< 0 and $\displaystyle |f(x)- f(a)|= |-x-(-a)|= |x-a|< \epsilon$ will be true as long as $\displaystyle \delta$ is less than the smaller of -a/2 and $\displaystyle \epsilon$.

Finally, if a= 0, |f(x)- f(a)|= | |x|- 0|= |x|. We can make that less than $\displaystyle \epsilon$ by taking $\displaystyle \delta= \epsilon$. - Feb 25th 2010, 08:18 PMCrazyCat87
Thank you HallsofIvy

- Feb 28th 2010, 05:27 AMCrazyCat87
- Feb 28th 2010, 05:41 AMCrazyCat87
Here's what I did, let me know if it's missing anything.

If $\displaystyle x>0$ then $\displaystyle f(x) = |x|=x$. As seen in some example, if $\displaystyle c\in\mathbb{R}$, then the limit of $\displaystyle f(x)$ as $\displaystyle x$ goes to $\displaystyle c$ equals $\displaystyle c$. Since $\displaystyle f(c) = c$, $\displaystyle f(x)$ is continuous at every point $\displaystyle c$.

Similar approach for $\displaystyle x<0$.... - Feb 28th 2010, 08:22 AMPlato
It is so easy. Learn the most useful inequality.

$\displaystyle \left( {\forall x~\&~y} \right)$ it is true that $\displaystyle \left| {\left| x \right| - \left| y \right|} \right| \leqslant \left| {x - y} \right|$.

Once we realize that, then we let $\displaystyle \delta = \varepsilon $.