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