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Using the definition of a limit, I am to show that the limit of $\sqrt{x}$ as $x$ approaches 1 is equal to one. This is what I have so far (I am stuck at the last line):

If $0<|x-1|<\delta $ then $|\sqrt{x}-1|<\varepsilon $

$\left|\frac{(\sqrt{x}-1)}{1}\frac{(\sqrt{x}+1)}{(\sqrt{x}+1)}\right|<\varepsilon $

$\frac{|x-1|}{\sqrt{x}+1}<\varepsilon $

$|x-1|<\varepsilon \sqrt{x}+\varepsilon $

I don't know if I've gone in the right direction. I know I need some $\varepsilon$ to equal $\delta$ and relate it back to $0<|x-1|<\delta$

Thank you very much.

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$|x-1|<\varepsilon \sqrt{x}+\varepsilon$

This is a great line. Since we know that $x \to 1$, why don't we just go ahead and say that we will never let $x$ get too far from $1$. Say... we will never consider $|x - 1| > \frac{1}{2}$. Then we know that $\varepsilon \sqrt x + \varepsilon < \dots$ and we can finish from there.

How does that look?