Prove continuity of sqrt(x) on (0,infinity)

[v41h4114]

Homework Statement

This is a problem from my Analysis exam review sheet.

Let L(x) = \sqrt{x}. Prove L is continuous on E = (0,\infty)

The Attempt at a Solution

The way we've been doing these proofs all semester is to let \epsilon > 0 be given, then assume \left| x -x_{0} \right| < \delta (which we figure out later) and x_{0} \in E

Then look at.
\left| L(x) - L(x_{0}) \right| = \left| \sqrt{x} - \sqrt{x_{0}}\right|
and try to get a \left( x -x_{0} \right) term which we can control, so that we can figure out a \delta which will allow us to get the entire thing less than \epsilon. So essentially I understand how to do the proof. My algebra skills are just really rusty and I can't figure out the long division or whatever I need to do to get a \left( x -x_{0} \right) term out of that.

 

[Office_Shredder]

One trick is to abuse inverse functions whose continuity you can deal with

If you let y^2=x and y_0^2 = x_0 then
|x-x_0| = |y^2-y_0^2| = |y_0-y||y_0+y|<\delta

Now you want to show that |y-y_0|<\epsilon if delta is small enough

 

[v41h4114]

Thanks for the quick response, I think I see where to go from here, but just to make sure.
Let \delta = \epsilon^{2}
We know \left| y + y_{0} \right| > \left| y - y_{0} \right| since y and y_{0} are positive.
and since \left| y + y_{0} \right| \left| y - y_{0} \right| < \delta = \epsilon^{2}
We can see that \left| y - y_{0} \right| < \sqrt{\delta} = \epsilon