Let xn(smaller n)>=0 for all n belong to N
a) if (xn)-->0, show that(sqrt[xn])-->0
b)if (xn)-->x,show that(sqrt[xn])-->x
Since x_n->0 as n->inf, for any e>0, we can find N>0, such that for all n>N, we have |x_n|<e^2. So |sqrt(x_n)|<e. This implies sqrt(x_n)->0 as n->inf.
These are proofs regarding limits.