Checking a Limit Proof

Question

I have to prove that $\displaystyle \lim_{x \to c} \sqrt{x}=\sqrt{c},\;c>0, x>0$

So I have to show that given any $\epsilon>0$, there exists a $\delta>0$ that for all x in the domain $0<|x-c|<\delta$ implies $|\sqrt{x}-\sqrt{c}|<\epsilon$

So I have

$|\sqrt{x}-\sqrt{c}|=|x-c|/(\sqrt{x}+\sqrt{c})<|x-c|/\sqrt{c}<\epsilon$,

So for any $\epsilon$>0, I let $\delta=\sqrt{c}\epsilon$,

So now

$|\sqrt{x}-\sqrt{c}|=|x-c|/(\sqrt{x}+\sqrt{c})<|x-c|/\sqrt{c}|<\delta/\sqrt{c}=\epsilon\sqrt{c}/\sqrt{c}=\epsilon$

Does this proof work?

Answer 1

I would use delta = minimum of your delta and c. This would force x to be positive and in the domain. (You wouldn't need to specify those conditions.)

Answer 2

As mentioned in the comments, your proof is correct.