What is the Proof for the Limit of Square Root Function?

[marco0009]

Homework Statement

Prove: \lim_{x \to a} \sqrt{x} = \sqrt{a}, if a>0

Homework Equations

|f(x)-L| < \epsilon
|x-a| < \delta

The Attempt at a Solution

|\sqrt{x}-\sqrt{a}| < \epsilon, when |x-a| < \delta
|x - 2\sqrt{x}\sqrt{a}-a| < \epsilon^2, when |x-a| < \delta

From here I don't know where to go. I don't see any obvious way to get delta into terms of epsilon.

 

[sutupidmath]

well |\sqrt{x}-\sqrt{a}| =|(\sqrt{x}-\sqrt{a})\frac{\sqrt{x}+\sqrt{a}}{\sqrt{x}+\sqrt{a}}| can you see what to do now?

 

[marco0009]

Yep. Thanks, it has been a long night.

 

[sutupidmath]

Yep. Thanks, it has been a long night.

It is 3 in the morning here, lol !