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.