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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,

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