Let $x_n$ be a sequence. Prove that $x_n \rightarrow 0 $ iff $ (x_n)^{2} \rightarrow 0 $
Attempt
Assume that $(x_n)^{2}$ converges to zero. So $| x_n|| x_n| \lt \epsilon'$ after some stage. Thus $| x_n|<\epsilon'/| x_n|$. Choose $\epsilon'$ to be $\epsilon | x_n|$. Is this an acceptable proof?
On
For the "if" part, we know that for all $\epsilon >0$, there exists a number $N(\epsilon)$ such that $|x_n|^2<\epsilon$ whenever $n>N$.
This means that for all $\epsilon>0$, $|x_n|<\epsilon$, whenever $n>M=N(\epsilon^2)$.
The same approach can be used to show the "only if part."
On
No, what you have to do is show that given $\varepsilon$ for one, you can always find $\varepsilon'$ for the other and vice versa.
One direction is easy: suppose $x_n \to 0$. Then by the definition of convergence there is an $N$ so that $\lvert x_n \rvert <1$ for $n>N$. Then you have $x_n^2<\lvert x_n \rvert $, because the left-hand side is the right-hand side multiplied by a positive number less than one. Now suppose you have $\varepsilon>0$, and $\lvert x_n \rvert < \varepsilon $ for $n>N'$, say. Then $x_n^2 < \lvert x_n \rvert <\varepsilon$ by the above, and hence for every $n>N'$, $x_n^2 < \varepsilon$. Since this holds for any $\varepsilon>0$, we then have $x_n^2 \to 0$.
Other way: suppose $x_n^2 \to 0$. Then given $\varepsilon>0$, there is an $N''$ such that $x_n^2<\varepsilon$ for all $n>N''$. It follows, by taking square roots of the inequality, that $$ -\sqrt{\varepsilon} < x_n < \sqrt{\varepsilon}, $$ or $\lvert x_n \rvert < \sqrt{\varepsilon} = \varepsilon'$, say. So we have that if $n>N''$, $\lvert x_n \rvert < \varepsilon' $, and for any such $\varepsilon'>0$, we can find a corresponding $N''$ so that $\lvert x_n \rvert < \varepsilon' $.
Alternative: show that $x_n^2$ is Cauchy iff $x_n$ is. $$ \lvert x_n-x_m \rvert^2 < \lvert x_n^2-x_m^2 \rvert = \lvert (x_n+x_m)(x_n-x_m) \rvert < 2A \lvert x_n-x_m \rvert, $$ the first being the triangle inequality, the second because $(x_n)$ converges, and hence is bounded in absolute value by $A$, say.
No, it's not acceptable. You have to prove that, from some point on, $|x_n|<\varepsilon$. And $\varepsilon$ must be fixed, you're not allowed to “choose one”. You are allowed to choose the place from where the inequality $|x_n|<\varepsilon$ holds.
Let $\varepsilon>0$. We know that there exists $m$ such that, for $n>m$, $|x_n|^2<\varepsilon^2$, which implies $|x_n|<\varepsilon$.
The other direction is similar. Suppose $\lim_{n\to\infty}x_n=0$. Fix $\varepsilon>0$. Then there is $m$ such that, for $n>m$, $|x_n|<\sqrt{\varepsilon}$. Then, for $n>m$, $|x_n^2|<\varepsilon$.