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### Section 2.10 : The Definition of the Limit

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

$\mathop {\lim }\limits_{x \to - 1} \left( {x + 7} \right) = 6$

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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,

$\left| {\left( {x + 7} \right) - 6} \right| < \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - \left( { - 1} \right)} \right| < \delta$

Or, with a little simplification this becomes,

$\left| {x + 1} \right| < \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x + 1} \right| < \delta$ Show Step 2

This problem is very similar to Problem 1 from this point on.

We need to determine a $$\delta$$ that will allow us to prove the statement in Step 1. However, because both inequalities involve exactly the same absolute value statement all we need to do is choose $$\delta = \varepsilon$$.

Show Step 3

So, let’s see if this works.

Start off by first assuming that $$\varepsilon > 0$$ is any number and choose $$\delta = \varepsilon$$. We can now assume that,

$0 < \left| {x - \left( { - 1} \right)} \right| < \delta = \varepsilon \hspace{0.5in} \Rightarrow \hspace{0.5in}0 < \left| {x + 1} \right| < \varepsilon$

This gives,

\begin{align*}\left| {\left( {x + 7} \right) - 6} \right| & = \left| {x + 1} \right| & & \hspace{0.25in}{\mbox{simplify things up a little}}\\ & < \varepsilon & & \hspace{0.25in}{\mbox{using the information we got by assuming }}\delta = \varepsilon \end{align*}

So, we’ve shown that,

$\left| {\left( {x + 7} \right) - 6} \right| < \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - \left( { - 1} \right)} \right| < \varepsilon$

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

$\mathop {\lim }\limits_{x \to - 1} \left( {x + 7} \right) = 6$