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 $\lim\limits_{x\rightarrow -1}(1-2x)=3$.