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Section 2.10 : The Definition of the Limit

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

\[\mathop {\lim }\limits_{x \to 3} x = 3\]

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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,

\[\left| {x - 3} \right| < \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - 3} \right| < \delta \]

This problem can look a little tricky since the two inequalities both involve \(\left| {x - 3} \right|\). Just keep in mind that the first one is really \(\left| {f\left( x \right) - L} \right| < \varepsilon \) where \(f\left( x \right) = x\) and \(L = 3\) and the second is really \(0 < \left| {x - a} \right| < \delta \) where \(a = 3\).

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In this case, despite the “trickiness” of the statement we need to prove in Step 1, this is really a very simple problem.

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 so 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

\[0 < \left| {x - 3} \right| < \delta = \varepsilon \hspace{0.25in} \Rightarrow \hspace{0.5in}0 < \left| {x - 3} \right| < \varepsilon \]

However, if we just look at the right portion of the double inequality we see that this assumption tells us that,

\[\left| {x - 3} \right| < \varepsilon \]

which is exactly what we needed to show give our choice of \(\delta \).

Therefore, according to the definition of the limit we have just proved that,

\[\mathop {\lim }\limits_{x \to 3} x = 3\]