Epsilon Delta Proof with $\frac{\sqrt x}{\sqrt x+1}$

Question

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

Answer 1

\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.