Prove: $\lim\limits_{x\rightarrow 2}x^{2}=4$.

Solution Given any $\epsilon >0$, we must show there is a number $\delta >0$ so that $$\begin{equation*} \hbox{whenever }\quad 0 \lt \vert x-2\vert \lt \delta \qquad \hbox{ then } \vert x^{2}-4\vert \lt \epsilon \end{equation*}$$

To establish a connection between $\vert x^{2}-4\vert$ and $\vert x-2\vert$, we write $\left\vert x^{2}-4\right\vert$ as $$\begin{equation*} \left\vert x^{2}-4\right\vert =\left\vert \,(x+2)(x-2)\right\vert =\left\vert x+2\right\vert \cdot \vert x-2\vert \end{equation*}$$

Now, if we can find a number $K$ for which $\left\vert x+2\right\vert \lt K$, then we can choose $\delta =\dfrac{\epsilon }{K}$. To find $K$, we restrict $x$ to some interval centered at $2$. For example, suppose the distance between $x$ and $2$ is less than $1$. Then $$\begin{equation*} \begin{array}{rcl@{\qquad}l} \\[-5pc] &&\vert x-2\vert \lt 1 & \\[3pt] -1&\lt&x-2 \lt 1 & \\[3pt] 1&\lt&x \lt 3 & {\color{#0066A7}{\hbox{Simplify.}}} \\ 1+2&\lt&x+2 \lt 3+2 & {\color{#0066A7}{\hbox{Add 2 to each part. }}} \\ 3&\lt&x+2 \lt 5 & \end{array} \end{equation*}$$

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In particular, we have $\vert x+2\vert \lt 5$. It follows that whenever $\vert x-2\vert \lt 1$, $$\begin{equation*} \vert x^{2}-4\vert =\left\vert x+2\right\vert \cdot \vert x-2\vert \lt 5\,\vert x-2\vert \hbox{ } \end{equation*}$$

If $\vert x-2\vert \lt \delta =\dfrac{\epsilon }{5}$, then $\left\vert \hbox{ }x^{2}-4\right\vert \lt 5\,\vert x-2\vert \lt 5\cdot \dfrac{\epsilon }{5}=\epsilon$, as desired.

But before choosing $\delta =\dfrac{\epsilon }{5}$, we must remember that there are two constraints on $\vert x-2\vert$. Namely, $$\begin{equation*} \vert x-2\vert \lt 1\qquad \hbox{and}\qquad \vert x-2\vert \lt \dfrac{\epsilon }{5} \end{equation*}$$

To ensure that both inequalities are satisfied, we select $\delta$ to be the smaller of the numbers 1 and $\dfrac{\epsilon }{5}$, abbreviated as $\delta =\min \left\{ 1,\dfrac{\epsilon }{5}\right\}$. Now, $$\hbox{whenever } \vert x-2\vert \lt \delta =\min \left\{ 1,\dfrac{\epsilon }{5}\right\} \hbox{ then } \vert x^{2}-4\vert \lt \epsilon$$

proving $\lim\limits_{x\rightarrow 2}x^{2}=4$.