Paul's Online Notes
Home / Calculus I / Limits / The Definition of the Limit
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### Section 2.10 : The Definition of the Limit

4. Use the definition of the limit to prove the following limit.

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

Show All Steps Hide All Steps

Start Solution

First, let’s just write out what we need to show.

Let $$\varepsilon > 0$$ be any number. We need to find a number $$\delta > 0$$ so that,

$\left| {{x^2} + 4x + 1 - \left( { - 2} \right)} \right| < \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - \left( { - 3} \right)} \right| < \delta$

Simplifying this a little gives,

$\left| {{x^2} + 4x + 3} \right| < \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x + 3} \right| < \delta$ Show Step 2

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

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,

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