Section 2.10 : The Definition of the Limit
1. Use the definition of the limit to prove the following limit.
limShow All Steps Hide All Steps
Start SolutionFirst, 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| < \deltaThis 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.
Show Step 2In 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 .
Show Step 3So, 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| < \varepsilonHowever, if we just look at the right portion of the double inequality we see that this assumption tells us that,
\left| {x - 3} \right| < \varepsilonwhich 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