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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
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1 Answer

  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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