Prove that the lim as x goes to a of sqrtx = sqrta

[apiwowar]

can i get some help on proving this limit?

 

[lurflurf]

start with
sqrt(x)-sqrt(a)=(x-a)/(sqrt(x)+sqrt(a))

 

[daviddoria]

I think it would be very helpful if you guys used latex

The question should be

Proove that $$\lim {x \rightarrow a} \sqrt{x} = \sqrt{a}$$

And the first reply should be

start with
$$\sqrt{x}-\sqrt{a}=\frac{x-a}{\sqrt{x}+\sqrt{a}}$$

For basic things like this, latex only takes about a minute to learn and it VERY much easier to read :)

 

[HallsofIvy]

In order to prove that "$\lim_{x\to a}\sqrt{x}= \sqrt{a}$, you need to prove that "Given $\epsilon> 0$, there exist $\delta> 0$ such that if $|x- a|< \delta$ then $|\sqrt{x}-\sqrt{a}|< \epsilon$".

Start from the second inequality and show (using the above hint) that you can find such a $\delta$.