Calculus I - The Definition of the Limit

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Section 2.10 : The Definition of the Limit

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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,

$\mathop {\lim }\limits_{x \to 3} x = 3$