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Show that if $x_n \to x$ then $\sqrt{x_n} \to \sqrt{x}$ I have been stuck on this for a while. I tried $$|\sqrt{x_n} - \sqrt{x}| = \frac{|x_n - x|}{|\sqrt{x_n} + \sqrt{x}|},$$ and then I at least can get the top to be as small as I want, so I have $$\frac{\epsilon}{|\sqrt{x+\epsilon} + \sqrt{x}|},$$ but I get stuck here at choosing the N, and I don't know if my first step in breaking down the absolute value is legitimate. Please help.

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If you separate the case $x=0$ from $x\ne 0$ you can prove the later with your method and the former directly. – KotelKanim Jan 15 '12 at 19:55
You might be interested in this post which asks how to prove that $x^{1/n}$ is continuous for all $n$ and all $x >0$. In particular, my answer there explains a similar idea as yours. – Srivatsan Jan 15 '12 at 20:19
  1. You want to treat the case $x=0$ separately.
  2. When $x\neq 0$, the identity you used is the way to go. Next use $\sqrt{x_n}+\sqrt{x}\geq \sqrt{x}$.
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Let $\epsilon > 0$. Since $(x_n)\rightarrow x$, then for some $N\in \mathbb{N}$, we have, $$|x_n-x|<\epsilon \sqrt{x} \hspace{10pt} \text{when}\hspace{10pt} n\geq N$$ This implies $$\frac{|x_n-x|}{\sqrt{x}}<\epsilon \hspace{10pt} \text{when}\hspace{10pt} n\geq N$$ Notice, \begin{align*} |\sqrt{x_n}-\sqrt{x}|&=|\sqrt{x_n}-\sqrt{x}|\left(\frac{\sqrt{x_n}+\sqrt{x}}{\sqrt{x_n}+\sqrt{x}}\right)\\ &=\frac{|x_n-x|}{\sqrt{x_n}+\sqrt{x}}\\ &\leq \frac{|x_n-x|}{\sqrt{x}} \end{align*} So, when $n\geq N$, $$|\sqrt{x_n}-\sqrt{x}|\leq\frac{|x_n-x|}{\sqrt{x}}<\epsilon $$ Hence, $(\sqrt{x_n})\rightarrow \sqrt{x}$.

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