So I've been trying to figure this out. The question is: If the limit x->infinity of Xn=Xo Show that, by definition, limit x->infinity sqrt(Xn)=sqrt(Xo) I'm pretty sure I need to use the epsilon definition. I worked on it with someone else and we think that what we have to show is the this: Want to show: For all e>0 there is an N>0 s.t. for all n>N, |sqrt(Xn) - sqrt(Xo)|<e I just don't know how to show this. Thanks!