\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.''

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$\displaystyle{
|f(x)-f(x_0)|
=\left|\sqrt x-\sqrt{x_0}\right|
=\left|{(\sqrt x-\sqrt{x_0})(\sqrt x+\sqrt{x_0})\over\sqrt x+\sqrt{x_0}}\right|
=\left|{x-x_0\over\sqrt x+\sqrt{x_0}}\right|
={|x-x_0|\over\sqrt x+\sqrt{x_0}}
}$

\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

$\displaystyle{
|f(x)-f(x_0)|
\le{|x-x_0|\over\sqrt{x_0}}
}$

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Step 3. We want to arrange for $|f(x)-f(x_0)|$ to be less than some epsilon
so we write

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$\displaystyle{
|f(x)-f(x_0)|
\le{|x-x_0|\over\sqrt{x_0}}<\epsilon
}$


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Step 4. Solving for $|x-x_0|$ we have

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$\displaystyle{
|x-x_0|<\epsilon\cdot\sqrt{x_0}
}$

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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

$\displaystyle{
|f(x)-f(0)|
=\left|\sqrt x-\sqrt{0}\right|
=\sqrt x<\epsilon
}$

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Solving for $x$ we have

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$\displaystyle{
x<\epsilon^2
}$

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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 $[0,\infty)$.

