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 $\lim\limits_{x\rightarrow c}\dfrac{1 }{x}=\dfrac{1}{c},$ $c\gt0$.