How would you show that the absolute value function is continuous at every point in

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- February 24th 2010, 07:43 PMCrazyCat87Absolute Value Function
How would you show that the absolute value function is continuous at every point in

- February 24th 2010, 07:44 PMDrexel28
- February 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, exist, and . 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 so that as long as , x> 0 and will be true as long as is less than the smaller of a/2 and .

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

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

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

If then . As seen in some example, if , then the limit of as goes to equals . Since , is continuous at every point .

Similar approach for .... - February 28th 2010, 08:22 AMPlato