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,

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