# Trouble understanding uniform continuity

I recognize intuitively what it means for a function to be continuous (i.e. no jumps or breaks in the function), but the concept of being uniformly continuous seems to be over my head.

I'm looking at the following function:

$f(x)=\frac{2^x}{x}$, $D=(0,5]$

How can I determine if this function is uniformly continuous or not?

• @Thomas I had a typo, I'm sorry. Thanks for your quick response – the_new_guy Jul 3 '16 at 18:39
• I'd say looking up the difference between continuity and uniform continuity in Google, and then maybe seeing if you can prove the ε-δ definition of uniform continuity holds true for this function on that domain. – gorzardfu Jul 3 '16 at 18:42
• @gorzardfu I've been reading the definitions over and over and just don't understand. I've even looked at plenty of examples and I just don't get it. – the_new_guy Jul 3 '16 at 18:44
• Watch this video: Continuous and Uniformly Continuous Functions – user209663 Jul 3 '16 at 19:05
• The distinction between continuity and uniform continuity can be thought of as one of local versus global. "Plain" continuity says that if you choose any point $a$ in the domain, and any tolerance $\epsilon > 0$, then you can choose a $\delta > 0$ dependent on both $a$ and $\epsilon$ such that if you choose $x$ within distance $\delta$ of $a$, then $f(a)$ and $f(x)$ will be within distance $\epsilon$ of each other. Uniform continuity is stronger, because it says we can choose $\delta$ dependent only on $\epsilon$, that is so this $\delta$ works for all $a$ in the domain. – AJY Jul 3 '16 at 19:16

A fast, intuitive, graphical approach to detect uniform continuity is to use the fact that a uniformly continuous real-valued function on a subset $A$ of the reals can be uniquely extended to a continuous function on the closure of $A$. In your case, the function blows up as $x$ tends to 0, so there is no way to continuously extend the function to $[0,5]$. Hence the function is not uniformly continuous.
• Note: the point is that to squeeze the graph of $f$ in an epsilon tube around $f(x)$ you must choose delta closer and closer to zero as you move $x$ closer and closer to zero. The function will however be uniformly continuous on $(a,5]$ for $0<a<5$. – symplectomorphic Jul 3 '16 at 20:01