Proof about sequence of non-negative real numbers 
How do I do this one? I don't know how to start.
 A: There is a slight technicality involved in proving $$\lim_{n \to \infty}x_{n}^{2} = L^{2}\Rightarrow \lim_{n \to \infty}x_{n} = L$$ This assertion is valid only when $x_{n} \geq 0$ after a certain point which implies that $L \geq 0$.
Moreover we need to differentiate the case $L^{2} > 0$ and $L^{2} = 0$. If $L^{2} > 0$ then we know that $L > 0$. Now for any given $\epsilon > 0$ we have a positive integer $N$ such that $$L^{2} - \epsilon < x_{n}^{2} < L^{2} + \epsilon$$ for all $n > N$. Since $L^{2} > 0$ it is possible to choose $\epsilon$ such that $0 < \epsilon < 3L^{2}/4 $ so that $$\frac{L^{2}}{4} < L^{2} - \epsilon < x_{n}^{2} < L^{2} + \epsilon$$ Since $x_{n}$ is non-negative it follows by taking square roots that $x_{n} > L/2$ and hence $x_{n} + L > 3L/2$. Now we can see that $$0 \leq |x_{n} - L| = \frac{|x_{n}^{2} - L^{2}|}{|x_{n} + L|} < \frac{2}{3L}|x_{n}^{2} - L^{2}| < \frac{2\epsilon}{3L}$$ for all $n > N$. It follows that $\lim_{n \to \infty}x_{n} = L$.
Next let $L^{2} = 0$ so that $L = 0$. We have $\lim_{n \to \infty}x_{n}^{2} = 0$ therefore for any $\epsilon > 0$ we have a positive integer $N$ such that $0 \leq x_{n}^{2} < \epsilon$ for all $n > N$. By taking square roots we get $0 \leq x_{n} < \sqrt{\epsilon}$ for all $n > N$. So $\lim_{n \to \infty}x_{n} = 0$.
A: How to start:


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*For the forward direction,  assume that $\lim x_n=L$.  Let $\epsilon>0$, and find some $N$ such that for all $n\ge N$, we must have $|x_n^2-L^2|\le \epsilon$.

*For the reverse direction, assume that $\lim x_n^2=L^2$.  Let $\epsilon>0$, and find some $N$ such that for all $n\ge N$, we must have $|x_n-L|\le \epsilon$.
A: I'll show if $\lim x_n=L$ then $\lim x_n^2=L^2$ using the delta epsilon definition of limits.
$\lim x_n=L$ implies for any $\varepsilon >0$, there exists an integer $N$ such that 
$|x_n-L|<\varepsilon $ for all $n>N$
Note $L-\varepsilon < x_n<L+\varepsilon$ so $2L-\varepsilon < x_n+L<2L+\varepsilon$, for all $n>N$.
Then 
$|x_n^2-L^2|=|(x_n+L)(x_n-L)|<|x_n+L||x_n-L|<(2L+\varepsilon )\varepsilon < \varepsilon^{\prime}$ for all $n>N$.
