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,

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