\beginsection 17.4 Prove that the function $\sqrt x$ is continuous on its domain $[0,\infty)$. {\it Hint:} Apply Example 5 in \S8. \medskip Step 1. We want to convert $|f(x)-f(x_0)|$ to an expression involving $|x-x_0|$. The trick from \S8 is ``irrationalize the denominator.'' \medskip $\displaystyle{ |f(x)-f(x_0)| =\left|\sqrt x-\sqrt{x_0}\right| =\left|{(\sqrt x-\sqrt{x_0})(\sqrt x+\sqrt{x_0})\over\sqrt x+\sqrt{x_0}}\right| =\left|{x-x_0\over\sqrt x+\sqrt{x_0}}\right| ={|x-x_0|\over\sqrt x+\sqrt{x_0}} }$ \medskip Step 2. We want to get the $\sqrt x$ out of the denominator so that we can solve for $|x-x_0|$. One thing we can use to our advantage is that we don't necessarily have to use the exact representation of $|f(x)-f(x_0)|$. All we really need to say is that $|f(x)-f(x_0)|$ is less than something. We can do that if we can find an expression that is less than $\sqrt x+\sqrt{x_0}$ since making the denominator smaller makes the right side of the equation larger. We notice that $\sqrt{x_0}\le\sqrt x+\sqrt{x_0}$ so we can write \medskip $\displaystyle{ |f(x)-f(x_0)| \le{|x-x_0|\over\sqrt{x_0}} }$ \medskip Step 3. We want to arrange for $|f(x)-f(x_0)|$ to be less than some epsilon so we write \medskip $\displaystyle{ |f(x)-f(x_0)| \le{|x-x_0|\over\sqrt{x_0}}<\epsilon }$ \medskip Step 4. Solving for $|x-x_0|$ we have \medskip $\displaystyle{ |x-x_0|<\epsilon\cdot\sqrt{x_0} }$ \medskip So if we choose $\delta=\epsilon\cdot\sqrt x_0$ then $|x-x_0|<\delta$ implies that $|f(x)-f(x_0)|<\epsilon$. \medskip Step 5. Note that the above fails for $x_0=0$. We have to show by another means that $\sqrt x$ is continuous at zero. First we write \medskip $\displaystyle{ |f(x)-f(0)| =\left|\sqrt x-\sqrt{0}\right| =\sqrt x<\epsilon }$ \medskip Solving for $x$ we have \medskip $\displaystyle{ x<\epsilon^2 }$ \medskip So if we choose $\delta=\epsilon^2$ then $|x-x_0|=x<\delta$ implies that $|f(x)-f(x_0)|=\sqrt x<\epsilon$. Note that we can say $|x-x_0|=x$ when $x_0=0$ because the domain we are using is $[0,\infty)$.