The 2.4 Precise Denition of a Limit
Math 1271, TA: Amy DeCelles
1. Overview
Denition of a Limit We say the limit of f (x) as x approaches a is L if the following condition is satised: For every number > 0 there is a number > 0 such that: if |x a| < then |f (x) L| < Parsing this denition Our intuitive understanding is that L is the limit (of f (x) as x a) if f (x) gets closer and closer to L as x gets closer and closer to a. Lets see how this matches up with the precise denition. First look at the expression |x a|. This is the distance between x and a. So if we make smaller and smaller, that means that x is getting closer and closer to a. Similarly, |f (x) L| is the distance between the y -values f (x) and L, so if gets smaller and smaller, that means that f (x) is getting closer and closer to L. So, just to make this clear: is a distance that species an x-range: how far away from a can x be? ... it must be within units of a. Well call this x-range a -neighborhood of a. And is a distance that species a y -range: how far away from L can f (x) be? ... it will be within units of L. This y -range well call an -neighborhood of L. So when we say that: if |x a| < then |f (x) L| < that is like saying: If x is close enough to a (namely in a -neighborhood of a) then f (x) is guaranteed to be close to L (namely in an -neighborhood of L). Ok, well then, what is the deal with the for every > 0 there is a > 0 part? This means that no matter how small you make the -neighborhood of L, you will always be able to nd a -neighborhood a, of that works, i.e. a -neighborhood small enough to guarantee that the y -values of the graph of f are in the -neighborhood of L. The point is that we can get f (x) innitely close to L, by just making the -neighborhood of a smaller and smaller.
2. Example
Problem: Prove, using the - denition of limit that:
x1
lim 5x 3 = 2
Solution: We need to show: For any > 0, there is a > 0 such that: if |x 1| < then |(5x 3) 2| <
Before we write our proof, we need to do some thinking. (This is like the prewriting you would do before writing a paper.) We treat like a xed number. We want to gure out what will work, given the we have. We start with the condition: |(5x 3) 2| < Simplifying, we get: |5x 5| < We factor out a 5: 5 |x 1| < So if we rewrite what we have to show, using the simplication we just did, we get: For any > 0, there is a > 0 such that: if |x 1| < then 5 |x 1| < Well, if |x 1| < then 5 |x 1| < 5 . So we just need to have 5 . So we will choose = 5 . Now that we have gured out what will work, we need to go back and write up an argument. (This is like writing a paper: we take the work we just did and arrange it nicely to construct an argument.) Proof: Given any
> 0, we can dene = 5 . Then: if|x 1| < : then |x 1| < 5 5
then 5 |x 1| < 5 then 5 |x 1| <
But 5 |x 1| = |(5x 3) 2|, so we have shown: if |x 1| < then |(5x 3) 2| < So we have shown: For any > 0, there is a > 0 such that: if |x 1| < then |(5x 3) 2| < and we are done!
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