Prove that if A_n converges to λ than square root of A_n converges to square root λ? Let an be a sequence of non-negative numbers converging to some number λ > 0. Prove that √an converges to √λ. (Hint: Note that multiplying and dividing dist(√an,√λ) by (√an + √λ) yields after a calculation that dist(√an,√λ) ≤ (1/√λ)dist(an,λ).
This is where I am: 
Let e be any real number greater than o. 
We want to prove that there exists an integer N such that:
All n > N | sqrt(λ) - sqrt(a_n) | < e. 
I don't really no where to move from here. The absolute values is tying my head in a knot, I don't know how to get rid of them in a correct way. Is this the type of proof I have to prove in cases? I notice when I rationalize it if the sum is negative or positive, I end up with the same thing. 

You can work "inside" them;
Since 
$$
\sqrt\lambda - \sqrt a_n = \frac{\lambda - a_n}{\sqrt\lambda + \sqrt a_n}
$$
they are equal in absolute value too! That is
$$
|\sqrt\lambda - \sqrt a_n| = \left|\frac{\lambda - a_n}{\sqrt\lambda + \sqrt a_n}\right|.
$$
Further $\sqrt \lambda + \sqrt a_n > \sqrt \lambda$ which is constant (so you can choose $N$ such that $|\lambda - a_n| < \sqrt\lambda\epsilon$).
