Epsilon-Delta Proof: Prove sqrt(x)=sqrt(a)

[tvguide123]

Homework Statement

Let a rep. any real number greater than 0
Prove that the limit as x->a of sqrt(x) = sqrt(a)

I have to prove the above equation using using an Epsilon-Delta proof but I am not sure how to start it off.

2. The attempt at a solution

I assumed that if 0<|x-a|<d
then |f(x) - f(a)|
= |sqrt(x) - sqrt(a)|

I am allowed to use basic manipulations of numbers that preserved the equation and also make helper assumption values for delta if needed as long as i account for them in my proof.

I've been stuck on this question for 3-1/2 hours now so I would really appreciate any help!

 

[daniel_i_l]

Try multiplying |sqrt(x) - sqrt(a)| by |sqrt(x) + sqrt(a)| /|sqrt(x) + sqrt(a)|

 

[zhentil]

Given epsilon>0, let delta = epsilon*sqrt(a), and remember that sqrt(x) + sqrt(a) >= sqrt(a) if x>=0.

 

[tvguide123]

Ah thanks a bunch guys, I couldn't figure out the first step for so long!

cheers :)