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I just really need to make sure that I am understanding the process for doing these.

Scratch work: We have $|\sqrt{x}-\sqrt{4}| = |\sqrt{x}-2|= |\frac{x-4}{\sqrt{x+2}}|= \frac{|x-4|}{|\sqrt{x}+2|}$. Here we suppose, $|x-4|<\delta<1 \Rightarrow \frac{|x-4|}{|\sqrt{x}+2|}<\frac{\delta}{2}.$ So we're done if we take, $\delta$=$min${$1,2\epsilon$}? In the previous steps I multiplied by the conjugate.

Mr.Fry
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One error crept in: $\sqrt x+2<5$ would make $\frac1{\sqrt x+2}>\frac15$ instead of $<\frac15$. However, we have $\frac 1{\sqrt x+2}\le \frac12$ right away without ado. So you can take $\delta=\min\{2\epsilon,4\}$. (You need to bound $\delta$ merely to have $x\ge0$ and $\sqrt x$ defined, but this is not strictly required)

Hagen von Eitzen
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  • Where did the 4 come from sorry? – Mr.Fry Sep 13 '13 at 06:06
  • @Crypto More generally, to show continuity of $x\mapsto\sqrt x$ at $x_0>0$, you can pick $\delta=\min{\sqrt{x_0}\epsilon, x_0}$. We shold ensure $\delta<x_0$ so that $\sqrt x$ is defined in the first place. Then again, the definition of continuity asks only for $\delta>0$ such that for all $x$ with $|x-x_0|<\delta$ for which $f(x)$ is defined - so this restriction is not really needed, you can take $\delta=\sqrt{x_0}\epsilon$. – Hagen von Eitzen Sep 13 '13 at 06:50