Originally Posted by
HallsofIvy 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 $\displaystyle \delta= \epsilon$.