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

\[\left| {{x^2} + 4x + 3} \right| = \left| {\left( {x + 1} \right)\left( {x + 3} \right)} \right| = \left| {x + 1} \right|\left| {x + 3} \right| < \varepsilon \]

We have the \(\left| {x + 3} \right|\) we expect to see but we also have an \(\left| {x + 1} \right|\) that we’ll need to deal with.

Show Step 3

To deal with the \(\left| {x + 1} \right|\) let’s first assume that

\[\left| {x + 3} \right| < 1\]

As we noted in a similar example in the notes for this section this is a legitimate assumption because the limit is \(x \to - 3\) and so \(x\)’s will be getting very close to -3. Therefore, provided \(x\) is close enough to -3 we will have \(\left| {x + 3} \right| < 1\).

Starting with this assumption we get that,

\[ - 1 < x + 3 < 1\hspace{0.25in} \to \hspace{0.25in}\,\, - 4 < x < - 2\] If we now add 1 to all parts of this inequality we get,

\[ - 3 < x + 1 < - 1\]

Noticing that -1 < 0 we can see that we then also know that \(x + 1 < 0\) and so provided \(\left| {x + 3} \right| < 1\) we will have \(\left| {x + 1} \right| = - \left( {x + 1} \right)\). Also, from the inequality above we see that,

\[1 < - \left( {x + 1} \right) < 3\]

All this means is that, provided \(\left| {x + 3} \right| < 1\), we will also have,

\[\left| {x + 1} \right| = - \left( {x + 1} \right) < 3\hspace{0.5in} \to \hspace{0.5in}\left| {x + 1} \right| < 3\]

This in turn means that we have,

\[\left| {x + 1} \right|\left| {x + 3} \right| < 3\left| {x + 3} \right|\hspace{0.25in}\,\,\,\,\,\,{\mbox{because }}\left| {x + 1} \right| < 3\]

Therefore, if we were to further assume, for some reason, that we wanted \(3\left| {x + 3} \right| < \varepsilon \) this would tell us that,

\[\left| {x + 3} \right| < \frac{\varepsilon }{3}\] 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

\[\delta = \min \left\{ {1,\frac{\varepsilon }{3}} \right\}\]

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

\[\delta \le 1\hspace{0.5in}{\mbox{AND}}\hspace{0.5in}\delta \le \frac{\varepsilon }{3}\]

Now assume that \(0 < \left| {x + 3} \right| < \delta = \min \left\{ {1,\frac{\varepsilon }{3}} \right\}\).

Show Step 5

So, let’s see if this works.

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

Now, let’s do the following,

\[\begin{align*}\left| {{x^2} + 4x + 3} \right| & = \left| {x + 1} \right|\left| {x + 3} \right| & & \hspace{0.25in}{\mbox{factoring}}\\ & < 3\left| {x + 3} \right| & & \hspace{0.25in}{\mbox{because we know }}\left| {x + 1} \right| < 3\\ & < 3\left( {\frac{\varepsilon }{3}} \right) & & \hspace{0.25in}\,{\mbox{because we know }}\left| {x + 3} \right| < \frac{\varepsilon }{3}\\ & = \varepsilon & & \end{align*}\]

So, we’ve shown that,

\[\left| {{x^2} + 4x + 3} \right| < \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x + 3} \right| < \min \left\{ {1,\frac{\varepsilon }{3}} \right\}\]

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

\[\mathop {\lim }\limits_{x \to - 3} \left( {{x^2} + 4x + 1} \right) = - 2\]