Introduction to the Calculus, Fall 2011 Help on Writing Epsilon Delta Proofs

Math 111, Introduction to the Calculus, Fall 2011 Help on writing epsilon/delta proofs The ‘Writing Mathematics’ document has some more general advice so please look at that as well. But there are some specific things to remember when using the epsilon/delta definition of a limit. 1. Do the scratch work and formal proof separately and label them as such. You should turn in the scratch work as well so that I know how you are doing the problem, but please separate it and mark it as scratch work. It is the formal proof that you will get credit for. If you mess up the proof, I might still give you some partial credit if the scratch work shows you have some idea of what is going on. 2. The formal proof needs to be explained using words. It is not enough just to write down a sequence of equations. The epsilon/delta notion of a proof is a complicated idea and you need to explain what you are doing. This means writing in complete sentences. You can of course use equations as part of your sentences, but, especially in these sorts of questions, the equations by themselves are not enough. See the examples below to see the difference. 3. Every limit proof has the same structure. Your proof should always look like this: Proof. Take any  > 0. Let δ = ? > 0. Then, for any x such that 0 < |x − a| < δ, we have .. . |f (x) − L| < . You can slightly different words to phrase it, but this is how it should look. Notice that this structure reflects the definition of the limit: lim f (x) = L means x→a for any  > 0, there is a δ > 0 such that: for any x with 0 < |x − a| < δ, we have |f (x) − L| <  4. Your choice of δ will almost always depend on , but is NOT allowed to depend on x. The idea is that for every  > 0, there is a δ > 0 that works, but it is usually a different δ for each . Therefore, when you pick a δ, it is likely to depend on , that is, the formula for δ involves . On the other hand, the values of x that we look at in the proof depend on δ. (They are all the values in the range a − δ < x < a + δ. This means it doesn’t make sense for δ to depend on x, because x is chosen after δ. 5. Showing that a limit is not equal to something is similar but sort of reversed. First you should make sure you understand, and know how to write proofs that a limit IS equal to something. To show that limx→a f (x) is NOT equal to L, we have to show that it is NOT true that: for any  > 0, there is a δ > 0 such that: for any x with 0 < |x − a| < δ, we have |f (x) − L| < . Sometimes it is hard to see what you have to do to show something is NOT true. In this case, you have to show that: there IS an  > 0, such that for ANY δ > 0, there is an x with 0 < |x − a| < δ and |f (x) − L| ≥ . To write up a proof like this, the structure should therefore be: Proof. Let  = ? > 0. Then take any δ > 0 and let x = ? . Then it is true that 0 < |x − a| < δ but, because of .. . it is NOT true that |f (x) − L| < . In this proof, you choose a particular number for , like 1 or 0.1. You also choose an x, but this is allowed to depend on both δ and . EXAMPLE 1 Prove that: lim (−2x + 5) = 3. x→1 BAD ANSWER: (This answer has the right elements but they are not in the right order and not explained at all.) Proof. |(−2x + 5) − 3| <  | − 2x + 2| <  2|x − 1| <  |x − 1| < /2 GOOD ANSWER: Scratch work: |(−2x + 5) − 3| = | − 2x + 2| = 2|x − 1| because |x − 1| = |1 − x|. (Math sucks.) 2|x − 1| < , |x − 1| < /2. So δ = /2. (I hate calculus.) (I can write whatever I want here because it’s only scratch work and Professor Ching will not read it.) End of scratch work. Proof. Take any  > 0 and let δ = /2 > 0. Then, if x is such that 0 < |x − 1| < δ we have |x − 1| < /2 and so 2|x − 1| <  so |2x − 2| <  so | − 2x + 2| <  so |(2x − 5) − (−3)| <  which is what we needed to show. EXAMPLE 2 Let f (x) be the function ( 1 if x is irrational; f (x) = 0 if x is rational. Prove that lim f (x) 6= 1. x→2 GOOD ANSWER: Scratch work: We want to show that as x gets close to 2, it is NOT true that all values of f(x) get close to 1. Well, the problem is that when x is rational and close to 2, the value of f (x) is not close to 1. In fact, those values are a distance 1 away from 1. So take  to be less than 1, say 1/2. Now take any δ > 0. We want to find an x that is within δ of 2 but such that f (x) is not within 1/2 if 1. So we want x to be a rational number that is within δ of 2. e.g. If δ = 0.0000032215, then we could take x = 2.000003, and so on. To explain this we could just say that there is always a positive rational number smaller than any given positive number. More explicitly, we could let x be what you get when you take the first non-zero digit in the decimal expansion of δ and delete the remaining digits, divide by 10 (just in case δ only had the one nonzero digit), then add 2. (End of scratch work.) Proof. Let  = 1/2 and take any δ > 0 Let x be the rational number obtained by x=2+ d 10 where d is given by taking the first non-zero digit in the decimal expansion of δ and deleting the rest. Then, since d 6= 0, we have x 6= 2, so 0 < |x − 2|. Also, since d ≤ δ, we have |x − 2| < δ. Altogether then 0 < |x − 2| < δ. But, because x is rational, we have f (x) = 0 and so |f (x) − 1| = 1 ≥ .