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
Solution Preview
(a) proof:
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.
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