- #1

Terrell

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- 26

## Homework Statement

Show that ##f(x)=\frac{1}{x^2}## is not uniformly continuous at ##(0,\infty)##.

## Homework Equations

N/A

## The Attempt at a Solution

Given ##\epsilon=1##. We want to show that we can compute for ##x## and ##y## such that ##\vert x-y\vert\lt\delta## and at the same time ##\vert f(x)-f(y)\vert\gt 1##. Note that

\begin{align}

\vert x-y\vert\lt\delta &\Leftrightarrow y-\delta\lt x \lt y+\delta

\end{align}

So let ##x=y+\delta/2##. Now to link the desired conditions on ##x## and ##y##, we compute for ##\vert \frac{1}{x^2}-\frac{1}{y^2}\vert \gt 1 \Leftrightarrow \vert \frac{1}{(y+\frac{\delta}{2})^2}-\frac{1}{y^2}\vert \gt 1 \Leftrightarrow \vert \frac{y\delta+\frac{\delta^2}{4}}{y^4+y^3\delta+(\frac{\delta}{2})^2}\vert \gt 1##.

Since ##\lim_{y\to 0}(\frac{y\delta+\frac{\delta^2}{4}}{y^4+y^3\delta+(\frac{\delta}{2})^2})=\infty##, since ##\delta## is fixed, then we can find ##x## and ##y## such that ##\vert f(x)-f(y)\vert \gt 1##.