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If the limit of sequence is zero then so is the limit of square roots

Solution: Let $\epsilon>0$. Since $\lim s_n=0$, there exists $N>0$ such that
|s_n-0|=s_n < \epsilon^2
$$ for all $n>N$. Therefore $n>N$ implies that
$$ for all $n>N$ as desired. Thus $\lim s_n=0$.

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