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

Prove \(\lim\limits_{x\rightarrow c}\dfrac{1}{x}=\dfrac{1}{c}\), where \(c\gt0\).

Solution The domain of \(f(x)=\dfrac{1}{x}\) is \(\{x|x\neq 0\} \).

For any \(\epsilon \gt0\), we need to find a positive number \(\delta \) so that whenever \(0 \lt \vert x-c\vert \lt \delta \), then \(\left\vert \dfrac{1}{x}-\dfrac{1}{c}\right\vert \lt \epsilon \). For \(x\neq 0\), and \(c\gt0\), we have \begin{equation*} \left\vert \frac{1}{x}-\frac{1}{c}\right\vert =\left\vert \frac{c-x}{xc} \right\vert =\frac{\vert c-x\vert }{\vert x\vert \cdot \vert c\vert }=\dfrac{|x-c|}{c\vert x\vert} \end{equation*}

The idea is to find a connection between \begin{equation*} \vert x-c\vert\qquad \hbox{ and }\qquad \dfrac{|x-c|}{c\vert x\vert } \end{equation*}

We proceed as in Example 3. Since we are interested in \(x\) near \(c\), we restrict \(x\) to a small interval around \(c\), say, \(\vert x-c\vert \lt \dfrac{c}{2}\). Then, \begin{equation*} \begin{array}{rcl@{\qquad}l} -\dfrac{c}{2}&\lt&x-c \lt \dfrac{c}{2}\\ \dfrac{c}{2}&\lt&x \lt \dfrac{3c}{2} & {\color{#0066A7}{\hbox{Add \(c\) to each expression.}}} \end{array} \end{equation*}

Since \(c \gt 0,\) then \(x\gt\dfrac{c}{2}\gt0,\) and \(\dfrac{1}{x} \lt \dfrac{2}{c}\). Now \begin{equation*} \left\vert \dfrac{1}{x}-\dfrac{1}{c}\right\vert =\dfrac{|x-c|}{c|x|} \lt \dfrac{2 }{c^{2}}\cdot |x-c|\qquad {\color{#0066A7}{\hbox{Substitute \(\dfrac{1}{\vert x\vert } \lt \dfrac{2}{c}\).}}} \end{equation*}

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We can make \(\left\vert \dfrac{1}{x}-\dfrac{1}{c}\right\vert \lt\epsilon \) by choosing \(\delta =\dfrac{c^{2}}{2}\epsilon .\) Then \begin{equation*} \hbox{whenever }\vert x-c\vert \lt\delta =\dfrac{c^{2}}{2} \epsilon ,\hbox{ we have }\left\vert \dfrac{1}{x}-\dfrac{1}{c} \right\vert \lt \dfrac{2}{c^2} \cdot |x-c| \lt \dfrac{2}{c^2}\cdot \left(\dfrac{c^2}{2}\cdot \epsilon\right)=\epsilon \end{equation*}

But remember, there are two restrictions on \(|x-c|\). \begin{equation*} \vert x-c\vert \lt\dfrac{c}{2}\qquad \hbox{ and }\qquad \vert x-c\vert \lt\dfrac{c^{2}}{2}\cdot \epsilon \end{equation*}

So, given any \(\epsilon \gt0\), we choose \(\delta =\min \! \left(\! \dfrac{c}{2},\dfrac{c^{2}}{2}\cdot \epsilon\!\! \right) \). Then whenever \(0\lt\vert x-c\vert \lt\delta \), we have \(\left\vert \dfrac{1}{x}-\dfrac{1}{c} \right\vert \lt\epsilon .\) This proves \(\lim\limits_{x\rightarrow c}\dfrac{1 }{x}=\dfrac{1}{c},\) \(c\gt0\).