Using the \(\epsilon \)-\(\delta\) Definition of a Limit

Use the \(\epsilon \)-\(\delta \) definition of a limit to prove \( \lim\limits_{x\rightarrow -1} (1-2x) =3\).

Solution Given any \(\epsilon \gt0\), we must show there is a number \(\delta \gt0\) so that \begin{equation*} \hbox{whenever }\quad 0 \lt \vert x-(-1) \vert \lt \delta \qquad \hbox{then }\vert ( 1-2x) -3\vert \lt \epsilon \end{equation*}

The idea is to find a connection between \(\vert x-(-1)\vert =|x+1|\ \) and \(\vert ( 1-2x) -3\vert \). Since \begin{equation*} \vert (1-2x)-3\vert =\vert -2x-2\vert = \vert -2(x+1)\vert = \vert -2\vert \cdot \vert x+1\vert =2\vert x+1\vert \end{equation*}

we see that for any \(\epsilon >0,\) \begin{equation*} \hbox{whenever}\quad \vert x-(-1)\vert =\vert x+1\vert \lt \dfrac{\epsilon }{2}\qquad \hbox{ then }\vert (1-2x)-3\vert \lt \epsilon \hbox{ } \end{equation*}

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That is, given any \(\epsilon \gt 0\) there is a \(\delta,\delta =\dfrac{\epsilon}{2}\), so that whenever \(0 \lt \vert x-(-1)\vert \lt \delta\), we have \(\vert (1-2x)-3\vert \lt \epsilon \). This proves that \(\lim\limits_{x\rightarrow -1}(1-2x)=3\).