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Linear functions

Here we want f(x) = x for example. We want to say $\lim_{x\rightarrow \pi}
f(x) = \pi$. So for every t > 0 we have to produce mt > 0 so that if $0 < \vert x-\pi\vert < m_t$ then $\vert f(x) - \pi \vert < t$. This is not hard either:

\begin{displaymath}
\vert f(x) - \pi\vert = \vert x - \pi\vert,\end{displaymath}

so mt could be any positive real number less than or equal to t. We can take mt = t.

Now look at something not quite so obvious: Here we want f(x) = 2x for example. We want to say $\lim_{x\rightarrow \pi}
f(x) = 2\pi$. So for every t > 0 we have to produce mt > 0 so that if $0 < \vert x-\pi\vert < m_t$ then $\vert f(x) - \pi \vert < t$.

\begin{displaymath}
\vert f(x) -2\pi\vert = \vert 2x - 2\pi\vert = 2\vert x-\pi\vert.\end{displaymath}

Now mt = t does not work, since f magnifies everything by a factor of 2. Thus we should compensate by taking mt any positive number less than or equal to half of t. We can take mt = t/2.

David G Radcliffe
8/18/1998