Calculus I - The Definition of the Limit

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Section 2.10 : The Definition of the Limit

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

lim x \to - 1 (x + 7) = 6

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

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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,

| (x + 7) - 6 | < ε whenever 0 < | x - (- 1) | < δ

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

Or, with a little simplification this becomes,

| x + 1 | < ε whenever 0 < | x + 1 | < δ

$\left| {x + 1} \right| < \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x + 1} \right| < \delta$ 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 $\delta = \varepsilon$ .

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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 < | x - (- 1) | < δ = ε \Rightarrow 0 < | x + 1 | < ε

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

This gives,

\begin{matrix} | (x + 7) - 6 | & = | x + 1 | & simplify things up a little < ε & using the information we got by assuming δ = ε \end{matrix}

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

So, we’ve shown that,

| (x + 7) - 6 | < ε whenever 0 < | x - (- 1) | < ε

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

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

lim x \to - 1 (x + 7) = 6

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