calculus - Suppose that $(s_n)$ converges to s. Prove that $(s_n^2)$ converges to $s^2$

I am really struggling with my proofs class, I don't really understand how to prove a statement like this, or what the epsilon is standing for..

Suppose that $(s_n)$ converges to s. Prove that $(s_n^2)$ converges to $s^2$ directly without using the fact that $lim(s_nt_n)=st$

Suppose that $(s_n)$ converges to s. Then, since $(s_n)$ is convergent, there exists an $M_1$ such that $|S_n|0$, there exists N such that n>N implies that $|s_n-s|<ε/M$. Thus for n>N, we have $|s_n^2-s^2|=|s_n-s|*|s_n+s|<(ε/M)(M)=ε$. Hence, $s_n^2$ converges to $s^2$.

Answer

Case (i): $s \neq 0$. Since $s_n \to s$, there exists $M$ such that for all $n>M$, we have $\vert s_n - s \vert < s/2$, i.e., $s/2 < s_n < 3s/2$. Again, since $s_n \to s$, given an $\epsilon > 0$, there exists $N$, such that $\vert s_n - s \vert < \dfrac{2\epsilon}{5s}$. Hence, choosing $K = \max(M,N)$, we get that
$$\vert s_n^2 - s^2 \vert = \vert s_n-s \vert \vert s_n + s \vert < \dfrac{2\epsilon}{5s} \left(\dfrac{3s}2 + s\right) = \epsilon$$

Case (ii): $s=0$. Since $s_n \to 0$, given any $\epsilon>0$, there exists $M$ such that $\vert s_n \vert < \min(\epsilon,1)$ for all $n>M$. Hence, $$\vert s_n^2 \vert < \min(\epsilon,1)^2 \leq \epsilon$$