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The question presented is

Use the $\epsilon -\delta$ definition of continuity to show that $$ f(x) = \frac{\sqrt{x}}{\sqrt{x}+1} $$ is continuous on $[0, \infty) $.

So my initial plan was to prove root x was continuous at the interval which I could do decently easily and then use the ratio test to prove the whole thing was continuous. My friend said that wouldn't work and I didn't quite get why. The prof also mentioned that I should maybe use the fact that $x_1 - x_2 = (\sqrt{x_1} - (\sqrt{x_2})(\sqrt{x_1} + (\sqrt{x_2}) $ and I'm not totally sure what to do with that.

Any help is super appreciated

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  • $\begingroup$ You want to find a $\delta$ such that whenever $|x-0| \lt \delta$ then $|f(x) -f(0)| = |\sqrt{x}| \lt \varepsilon$. Set $x_2 = 0$ and $x_1 = x$ in the hint they gave you. $\endgroup$ – mate89 Dec 4 '17 at 5:15
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\begin{align*} f(x)=1-\dfrac{1}{\sqrt{x}+1}, \end{align*} so for fixed $y\in(0,\infty)$, given $\epsilon>0$, for all $x$ with $|x-y|<\min\{y/2,\epsilon\}$, \begin{align*} |f(x)-f(y)|&=\left|\dfrac{\sqrt{x}-\sqrt{y}}{(\sqrt{x}+1)(\sqrt{y}+1)}\right|\\ &=\dfrac{|x-y|}{(\sqrt{x}+1)(\sqrt{y}+1)(\sqrt{x}+\sqrt{y})}\\ &\leq\dfrac{|x-y|}{\sqrt{x}+\sqrt{y}}\\ &\leq\dfrac{|x-y|}{\sqrt{y/2}+\sqrt{y}}. \end{align*}

For the case that $y=0$ is easier.

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  • $\begingroup$ oh man. I didnt think of using that type of alternate value. I'll have to look through and this and write up something that works for me, but thank you so much. I'll need to still look at the continuity at 0 because the interval is [0, $\infty$) but thanks for the help $\endgroup$ – Jebus Christo Dec 4 '17 at 5:26

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