Epsilon Delta proofs Before reading this guide, please go over inequalities (if needed). Eample Prove lim(4 3) = 5 2 First we have to know what the definition of a limit is: i.e rigorous way of saying lim a f() = L For every ε > 0 ( ε > 0), there eists a δ > 0 ( δ > 0) such that f() L < ε whenever 0 < a < δ A bit on the symbols: The symbol means for all or for every. The symbol means there eists The Greek letter ε is epsilon and the Greek letter δ is delta. The last sentence in the bo above means that when considering all the numbers that are δ units from a (those are the s) when the function f evaluates those s, f(), and subtracts L (the limit you are claiming), the difference is less than ε. The task at hand is given ε > 0, find a δ > 0. This δ is dependent of ε only; meaning our choice of δ relies only on ε and not. It is often the case in math that you work with the result your problem is asking you to prove. In this problem and subsequent problems, we will start with the part f() L < ε. We do this because it is easier to work with. We can use algebra/trigonometry much of the time to break things down to where they look like something we can work with. I will replace f() with 4 3 and L with 5. I then will use algebra to simplify the inequality. 4 3 5 < ε whenever 0 < 2 < δ 4 < ε 4 2 < ε
2 < ε 4 We arrive at a point where 2 < ε matches the part of the definition of 0 < a < δ, so 4 I would pick δ = ε 4. We made a very educated guess. The algebra we did above is often called the scratch work and often leads us to find the value of δ. We will use this method in this guide. What we need to do is actually prove this delta works!! Proof Given ε > 0 and let = ε 4. Consider 0 < 2 < ε 4 4 2 < ε 4 < ε (4 3) 5 < ε f() L < ε The square means that we are done with our proof. You will often see this square or the acronym Q.E.D to notify the reader that our proof is complete. It might have seemed like I worked backwards when working the proof out, which I did. Much of the time, epsilon delta proofs are written in that manner. The value of δ is not unique. Once, you have found a value for δ any smaller value of δ will also work because we are working with inequalities. For eample, if 2 < then 2 <. The values of δ = ε, ε, ε would also work with eample. We will use this idea for 2 5 00000 the net problem. Eample 2 Prove lim 3 (2 + + ) = 3 2 + + 3 < ε whenver 0 < 3 < δ
So we might want to choose δ = ε, not on. ε +4 2 + + 3 < ε 2 + 2 < ε ( + 4)( 3) < ε + 4 3 < ε 3 < ε +4. However, this is not correct at all because δ depends on So how do we remedy this situation? There are a couple options but I will use the option that I think is the most useful for other problems. What I will do is bound δ, meaning that δ will be less than or equal to some number say. So we will have δ. Another way to look at this is to say we are looking at the numbers that are less than unit from 3. I chose because it s a nice number to work with. It could have also been 5, 3.00725, etc.. Any number really. Now, why can we do this? What s the objection to restricting δ? If you were to say how about if my δ >? My answer would be choose δ = ; as long as one δ works, 2 then any δ smaller than that value will also work. Now, why am I doing this? If I can restrict the value of δ, then we can perhaps say something about the value of + 4 3. More specifically, + 4 some number Well if δ, then 3 < < 3 < 2 < < 4 In other words, all the numbers that are unit away from 3 are between 2 and 4. If 2 < < 4 6 < + 4 < then + 4 3 < 3 So if let s say 3 < ε then 3 < ε. My pick δ = ε But here is the catch. The value that I picked for δ is dependent that δ. How can I make sure that δ = ε is not greater than? Certainly, if I someone was to give me ε = 0,2 then δ = ε will be bigger than. So how do I guarantee a delta that will work?
We do this: Let δ = min(, ε ). The min means that we will be choosing the smaller value of and ε. For eample, min(,.5) =.5, min(,0) = this will guarantee that δ and it guarantees that the two inequalities 3 < ε and 3 < hold. Proof Given ε > 0, let δ = min(, ε ). Suppose 0 < 3 < δ, then 3 < and 3 < ε. If 3 <, then < 3 < and + 4 <. Using + 4 < and 3 < ε we have: + 4 3 < ε 2 + 2 < ε ( 2 + + ) 3 < ε A note about the validity of the inequality + 4 3 < ε ; I am using a property about inequalities which states that if a < b and c < d where a, b, c and d are positive real numbers, then ac < bd Eample 3 Find a value for δ given ε. + lim 2 = (This will not be a proof. I strongly recommend you do complete the proof of this limit) + 2 < ε whenever 0 < < δ < ε 2 2 < ε
Were back to the problem we had with our last eample. Again, I will restrict δ. Let s make the same restriction, δ. If <, this would imply 0 < < 2. Now notice how 2 is on the denominator. If 0 < < 2, what can I say about? Well 0 < 2 < 4 but all I can say is >. To 2 2 4 eplain this, numbers like.000,.00000 lie within the inequality 0 < 2 < 4. When we divide these numbers into, then we would get numbers like 0000, 000000, etc. Basically, we will grow without bound. Unfortunately, this will not help us because we want to bound the inequality. Note, that there is nothing wrong with restricting δ ; it s just not giving us a result that we could work with to bound the epression 2. To remedy the situation, δ. < this means < < 3 and < 2 < 3 2 2 2 2 and 3 < 2 c < b < a ) Hence, 2 < ( I am using an inequality that states that if a < b < c where a, b, c > 0 then < and if < ε, then δ = ε Just like our last eample, δ = min (, ε) Eample 4 Prove lim sin 0 = 0 sin 0 < ε whenever 0 < 0 < δ sin < ε sin < ε We know that sin(θ), then certainly sin θ. But because θ is arbitrary, sin
Hence sin, so if < ε then δ = ε Proof Given ε > 0, let δ = ε. Suppose 0 < < ε sin < ε = ε sin < ε sin 0 < ε The last eample showed that even though we had the epression, sin, we did not have to put any restriction on δ. The epression having a trigonometric function gave us a restriction for sin