Let’s start with a little simplification of the first inequality.

\[\left| {{x^2} - 4} \right| = \left| {\left( {x + 2} \right)\left( {x - 2} \right)} \right| = \left| {x + 2} \right|\left| {x - 2} \right| < \varepsilon \]

We have the \(\left| {x - 2} \right|\) we expect to see but we also have an \(\left| {x + 2} \right|\) that we’ll need to deal with.

Show Step 3

To deal with the \(\left| {x + 2} \right|\) let’s first assume that

\[\left| {x - 2} \right| < 1\]

As we noted in a similar example in the notes for this section this is a legitimate assumption because the limit is \(x \to 2\) and so \(x\)’s will be getting very close to 2. Therefore, provided \(x\) is close enough to 2 we will have \(\left| {x - 2} \right| < 1\).

Starting with this assumption we get that,

\[ - 1 < x - 2 < 1\hspace{0.25in} \to \hspace{0.25in}\,\,1 < x < 3\]

If we now add 2 to all parts of this inequality we get,

\[3 < x + 2 < 5\]

Noticing that 3 > 0 we can see that we then also know that \(x + 2 > 0\) and so provided \(\left| {x - 2} \right| < 1\) we will have \(x + 2 = \left| {x + 2} \right|\).

All this means is that, provided \(\left| {x - 2} \right| < 1\), we will also have,

\[\left| {x + 2} \right| = x + 2 < 5\hspace{0.5in} \to \hspace{0.5in}\left| {x + 2} \right| < 5\]

This in turn means that we have,

\[\left| {x + 2} \right|\left| {x - 2} \right| < 5\left| {x - 2} \right|\hspace{0.25in}\,\,\,\,\,\,{\mbox{because }}\left| {x + 2} \right| < 5\]

Therefore, if we were to further assume, for some reason, that we wanted \(5\left| {x - 2} \right| < \varepsilon \) this would tell us that,

\[\left| {x - 2} \right| < \frac{\varepsilon }{5}\] Show Step 4

Okay, even though it doesn’t seem like it we actually have enough to make a choice for \(\delta \).

Given any number \(\varepsilon > 0\) let’s chose

\[\delta = \min \left\{ {1,\frac{\varepsilon }{5}} \right\}\]

Again, this means that \(\delta \)will be the smaller of the two values which in turn means that,

\[\delta \le 1\hspace{0.5in}{\mbox{AND}}\hspace{0.5in}\delta \le \frac{\varepsilon }{5}\]

Now assume that \(0 < \left| {x - 2} \right| < \delta = \min \left\{ {1,\frac{\varepsilon }{5}} \right\}\).

Show Step 5

So, let’s see if this works.

Given the assumption \(0 < \left| {x - 2} \right| < \delta = \min \left\{ {1,\frac{\varepsilon }{5}} \right\}\) we know two things. First, we know that \(\left| {x - 2} \right| < \frac{\varepsilon }{5}\) . Second, we also know that \(\left| {x - 2} \right| < 1\) which in turn implies that \(\left| {x + 2} \right| < 5\) as we saw in Step 3.

Now, let’s do the following,

\[\begin{align*}\left| {{x^2} - 4} \right| & = \left| {x + 2} \right|\left| {x - 2} \right| & & \hspace{0.25in}{\mbox{factoring}}\\ & < 5\left| {x - 2} \right| & & \hspace{0.2in}\,{\mbox{because we know }}\left| {x + 2} \right| < 5\\ & < 5\left( {\frac{\varepsilon }{5}} \right) & & \hspace{0.25in}\,{\mbox{because we know }}\left| {x - 2} \right| < \frac{\varepsilon }{5}\\ & = \varepsilon & & \end{align*}\]

So, we’ve shown that,

\[\left| {{x^2} - 4} \right| < \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - 2} \right| < \min \left\{ {1,\frac{\varepsilon }{5}} \right\}\]

and so by the definition of the limit we have just proved that,

\[\mathop {\lim }\limits_{x \to 2} {x^2} = 4\]