Perhaps this is not what you're looking for, but sqrt is continuous, so the limit of f(x_n)=f(lim x_n). Or are you confused about epsilon delta?
This one is giving me such a hard time. I guess this is important to know because it is saying that the square root of the function translates to the square root of the limit. Similar to mutliplication and addition. The hint they give is to use the fact that if all convergent subsequences have the same limit, then the sequence is convergent. I need help on this one. Thanks.
Suppose <x_n> --->x for each natural number n, where x_n >= 0. Prove that <sqrt(x_n)> ---> to sqrt(x)
You can use the continuous inverse theorem on on to get continuity for the whole thing. This result follows from the preservation of interval theorem.
By the way, I think you need to follow the logic on the second proof provided here, because mine actually has a fault in it. I didn't even think about the fact that in which case that inequality isn't true... So yea, go with the mult/div by that expression to fix it. Sorry!