Applying the Limit Definition

  • #1
Hi. I'm a first-year calculus student and I'm fairly behind with my work. The transition is tough and when i read my textbook, I don't fully absorb everything. I thought I would post an example problem whose solution I do not follow completely, since it is fairly important in the scope of things.

I have provided a diagram for this problem and the solution provided by the author. If anyone would be able to explain to me in simpler terms how the conclusion was reached, I would greatly appreciate it.

EDIT: I have previewed my post and found some very weird issues when trying to use LaTeX, so I have removed the [tex [/tex] brackets, since everything displayed very improperly. I am not very experienced with LaTeX, so there may be something I am doing wrong, so if anyone could point that out, I will gladly fix up the presentation of my post.

But here is the problem:

http://i.imgur.com/nDyAv.jpg

For any number a, with 0 < a < 1, the function f approaches 0 at a. To prove this, consider any number \epsilon > 0. Let n be a natural number so large that \frac{1}{n} \leq \epsilon. Notice that the only numbers x for which |f(x) = 0| < \epsilon could be false are

\frac{1}{2}; \frac{1}{3}, \frac{2}{3}; \frac{1}{4}, \frac{3}{4}...; \frac{1}{n},...,\frac{n-1}{n}

(If a is rational, then a might be one of these numbers.) However many of these numbers there may be, there are, at any rate, only finitely many. Therefore, of all these numbers, one is closest to a; that is, |p/q - a| is smallest for one p/q among these numbers. (If a happens to be one of these numbers, then consider only the values |p/q - a| for p/q \neq a. This closest distance may be chosen as the \delta. For if 0 < |x-a| < \delta, then x is not one of

\frac{1}{2},...,\frac{n-1}{n} and therefore |f(x)-0| < \epsilon. This completes the proof. Note that our description of the delta which works for a given epsilon is completely adequate - there is no reason why we must give a formula for delta in terms of epsilon.
 

Answers and Replies

  • #2
36
0
For any number a, with 0 < a < 1, the function f approaches 0 at a. To prove this, consider any number [tex]\epsilon > 0[/tex]. Let n be a natural number so large that

[tex]\frac{1}{n} \leq \epsilon[/tex].

Notice that the only numbers x for which [tex]|f(x) = 0| < \epsilon[/tex] could be false are

[tex]\frac{1}{2}; \frac{1}{3}, \frac{2}{3}; \frac{1}{4}, \frac{3}{4}...; \frac{1}{n},...,\frac{n-1}{n}[/tex]

(If a is rational, then a might be one of these numbers.) However many of these numbers there may be, there are, at any rate, only finitely many. Therefore, of all these numbers, one is closest to a; that is, |p/q - a| is smallest for one p/q among these numbers. (If a happens to be one of these numbers, then consider only the values |p/q - a| for [tex]p/q \neq a[/tex]. This closest distance may be chosen as the [tex]\delta[/tex]. For if [tex]0 < |x-a| < \delta[/tex], then x is not one of

[tex]\frac{1}{2},...,\frac{n-1}{n}[/tex] and therefore [tex]|f(x)-0| < \epsilon[/tex].

This completes the proof. Note that our description of the delta which works for a given epsilon is completely adequate - there is no reason why we must give a formula for delta in terms of epsilon.

The basic idea is that you choose delta so that there is no fraction with denominator less than n within delta of a. You can then say that within delta of a, f is smaller than 1/n, except possibly at a itself.
 
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