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 $\left| y - y_{0} \right| < \sqrt{\delta} = \epsilon$