The Epsilon-Delta Limit Definition:

The Epsilon-Delta Limit Definition: A Few Examples Nick Rauh 1. Prove that lim x a x 2 = a 2. (Since we leave a arbitrary, this is the same as showing x 2 is continuous.) Proof: Let > 0. We wish to find δ > 0 such that for any x R, 0 < x a < δ implies x 2 a 2 <. We claim that the choice { } 2a + 1, 1 is an appropriate choice of δ. First note that 2a is always nonnegative, so 2a + 1 is always positive. This means 2a +1 > 0 for any a R, so we have { actually chosen δ > 0. Next we observe that if x a < 2a +1 },, 1 then x 2 a 2 = x + a x a (factoring) < ( 2a + 1) x a ( x + a < 2a + 1 if x a < 1) ( ) < ( 2a + 1) x a < 2a + 1 2a + 1 =. Thus, we see that this choice of δ forces x 2 a 2 < whenever 0 < x a < δ. It follows that lim x 2 = a 2. x a 1

Discussion Once we have a δ > 0 that we think could work, all we have to do is plug it in as we did above and verify. But how did we come up with our particular choice of δ > 0? We basically had to work backwards. We started with the idea that given > 0 we want δ > 0 such that for any x R, 0 < x a < δ implies x 2 a 2 <. We then had two inequalities to work with and somehow wanted to use them to come up with an expression for δ. We might first note that we can factor x 2 a 2 into x + a x a. We could then rewrite our task to be finding a δ0 such that x a < δ implies x + a x a <. The nice thing about this new task is that we have a common coefficient in both of our inequalities: x a < δ x + a x a < Perhaps this could be our in to combining the inequalities and getting an expression for δ. Our first step towards this would be isolating the common coefficient in our expression for : x a < x + a. We then note that if we were to choose δ / x + a, then it would certainly be the case that x a < δ x + a x a x + a <. However, δ > 0 would then depend on x. However, we want x a < δ. This means we actually want x to depend on δ! So we need to find a choice of δ > 0 such that δ does not depend on x. We know that choosing δ / x + a will get the job done. The question is how we can choose δ in this way. Well, we know that whenever we have one δ > 0 that works, any smaller value will also work. This means we can assume δ < 1. By choosing δ 1, we would have x a < 1 whenever x a < δ. This is the part where the reasoning gets tricky. If x a 1, this is the same as saying that x is no further away from a than a distance of 1. If x is no further away from a than a distance of 1, what 2

can we say about x + a? Well, we can say that x + a is no further away from 2a than a distance of 1. We can then write x+a < 2a +1. We then notice that we have just created a bound for x + a, which was the term in δ = / x + a that made this choice of δ dependent on x. We note that if we replace the denominator with 2a + 1, we get { } x a < 2a + 1, 1 < x + a. The reason that we have to take the minimum of /( 2a + 1) and 1 is because we had shown that x + a < 2a + 1 on the condition that x a < δ 1. If we had chosen to bound δ above by some other value say, δ < 20 then we would find that x + a < 2a + 20 on the condition that x a < δ 20, and thus choose { } 2a + 20, 20. We have thus constructed a δ > 0 that we think will work. All that is left for us to do is go back and make sure that for this δ > 0, 0 < x a < δ implies x 2 a 2 <, as we did in the proof. 3

2. Prove that lim x 2 x 3 = 8. (This proves that x 3 is continuous at x = 2.) Proof: Let > 0. We want δ > 0 such that for any x R, 0 < x 2 < δ forces x 3 8 <. We claim that { } 19, 1 gets the job done. To see that this, suppose x 2 < { 19, 1}. Then we must have x 3 8 = x 2 x 2 + 2x + 4 (factoring) < x 2 19 ( x 2 + 2x + 4 < 19 if x 2 < 1) < ( 19 19 x 2 < ) 19 =. Thus, we have that this δ forces x 3 8 < whenever 0 < x 2 < δ. It follows that lim x 3 = 8. x 2 4

Discussion Again, we have created a δ > 0 as if by magic. How would we go about finding such a δ? As last time, we were given > 0 and wanted δ > 0 such that for any x R, 0 < x 2 < δ implies x 3 8 <. We again had two inequalities to work with and somehow wanted to use them to create an expression for δ. Since we are again dealing with a polynomial, our gut instinct is to attempt to factor it. This gives x 3 8 = x 2 x 2 + 2x + 4, allowing us to rephrase our question. We want δ > 0 such that x 2 < δ implies x 2 x 2 + 2x + 4 <. Just like last time we might then consider δ = x 2 + 2x + 4, since x 2 < δ = x 2 + 2x + 4 x 2 x 2 + 2x + 4 <. However, this choice of δ is dependent on x, which is no good. We can save ourselves the same way we did last time, though. If we have a good choice of δ > 0, then any smaller choice will also work, so we can place the restriction δ 1. This would then mean that whenever x 2 < δ, then x 2 < 1. Unravelling this inequality, we have x 2 < 1 1 < x 2 < 1 1 < x < 3. (Here, is the mathematical symbol for these two statements are equivalent. ) Since x 2 + 2x + 4 is the part of our δ expression that is causing problems, we will try and use this restriction 1 < x < 3 to restrict x 2 + 2x + 4. We begin by noting that 1 < x < 3 implies the following two inequalities: 2 < 2x < 6 and 1 < x 2 < 9. Putting these together gives us the following: 3 < x 2 + 2x < 15 7 < x 2 + 2x + 4 < 19. Here comes the slightly tricky part in the reasoning. Since x 2 < 1, x can be no further away from 2 than a distance of 1. This means that x is positive. If x is positive, then so is x 2 + 2x + 4, so we must have x 2 + 2x + 4 = x 2 + 2x + 4 7 < x 2 + 2x + 4 < 19. 5

In this way we have bounded x 2 +2x+4. Going back to our previous argument, we note that as long as we choose δ / x 2 + 2x + 4, it should be a good choice of δ. Since x 2 +2x+4 < 19 when x 2 < 1, we might then choose { } 19, 1, since 19 > x 2 + 2x + 4 makes x 2 < δ 19 < x 2 + 2x + 4. To formulate a proof, it would then remain for us to show that this δ gives x 3 8 < whenever x 2 < δ, as we did in the proof above. 6

3. Prove that lim x 1 x + 2 = 3. (This shows x + 2 is continuous at x = 1.) Proof: Suppose we are given > 0. We then wish to find δ > 0 such that for any x R, 0 < x 1 < δ implies x + 2 3 <. We claim that the choice {, 1} is appropriate. To verify this, suppose x 1 < {, 1}. Then x 1 < < x 1 < (unravelling the absolute value) 3 < x + 2 < 3 + (adding 3 to all sides) 3 < x + 2 < 3 + ( x + 2 > 0 if x 1 < 1) < x + 2 3 < (subtracting 3 from all sides) x + 2 3 < (combining the inequalities) Thus, we see that this choice of δ forces x + 2 3 < whenever 0 < x 1 < δ. Thus, lim x + 2 = 3. x 1 7

Discussion Again, we found a δ > 0 that works, so let s see how we might have found it. Unlike last time, we have no polynomials to factor. We do, however, have an excessive number of absolute value symbols. When in doubt, unravel those things. We ll start (as before) with our inequality for : x + 2 3 < < x + 2 3 < 3 < x + 2 < 3 +. We now have an absolute value wedged between two values. If we unravel this thing, we actually end up with two sets of inequalities. The first type of inequality we have seen before: (3 + ) < x + 2 < 3 + 5 < x < 1 +. The second type is something new: x + 2 > 3 or x + 2 < (3 ). The tricky part in the reasoning is then as follows. We are of the mindset that is going to be arbitrarily small, so for the values of that are worth considering, (3 ) < 0. However, if we set the requirement δ 1, which is completely within our right, we would then have x 1 < 1, so x must be positive. If x > 0, then certainly x + 2 > 0. Thus, in the above inequality, the option x + 2 < (3 ) makes no sense, since it claims a positive number is less than a negative number. We must then choose x + 2 > 3 x > 1. Thus, we have the following two sets of inequalities: 5 < x < 1 + and x > 1. We can then try to combine them. We note that since 5 < 1 and 1 < x, we can combine them in the following way: 5 < 1 < x < 1 + 1 < x < 1 + 8 < x 1 <.

Since one of our bounds is the negation of the other, we may then fold this back into an absolute value inequality: x 1 <. It might then seem like is an appropriate choice of δ > 0. However, we have to remember that part of our reasoning involved forcing δ 1, so we must instead suggest {, 1}. To complete a proof, we would then check whether this choice of δ is such that x + 2 3 < whenever 0 < x 1 < δ, as we did in our proof above. 9

4. Prove that lim x 1 (4 + x 3x 3 ) = 2. (This proves that 4 + x 3x 3 is continuous at x = 1.) Proof: Suppose > 0. We want to find some δ > 0 such that for any x R, 0 < x 1 < δ forces (4 + x 3x 3 ) 2 <. We claim that { } 20, 1 would be an alright choice for δ. To check this, suppose x 2 < { 20, 1}. Then we see that (4 + x 3x 3 ) 2 = 3x 3 + x + 2 (simplification) = x 1 3x 2 + x + 2 (factoring) < x 1 20 ( 3x 2 + x + 2 < 20 if x 1 < 1) < ( 20 20 x 1 < ) 20 =. Thus, we have that this δ forces (4 + x 3x 3 ) 2 < whenever 0 < x 1 < δ. Thus, lim(4 + x 3x 3 ) = 2. x 1 10

Discussion Once again, it is time to discuss how we might find the δ that we did. We want to find δ > 0 such that (4 + x 3x 3 ) 2 < whenever 0 < x 1 < δ, so we are (as always) attempting to coerce something useful out of two inequalities. Since we are back to dealing with polynomials, our gut instinct is to factor. Our polynomial is even messier than the previous two we have looked at, but we argue that we can still use a factoring trick. We note that (4 + x 3x 3 ) 2 = 3x 3 + x + 2 = x 1 3x 2 3x 2. Now our problem is equivalent to wanting δ > 0 such that x 1 < δ implies x 1 3x 3 + x + 2 <. Our natural first guess for δ is δ = 3x 2 3x 2, since x 1 < δ = 3x 2 3x 2 x 1 3x 2 3x 2 <. However, just as with our two previous polynomial examples, this δ is dependent on x, so we need a different δ. Again, if we have a δ > 0 that works, then any smaller choice will also work, so place the restriction δ 1. This gives that whenever x 1 < δ, then x 1 < 1. Unravelling this inequality, x 1 < 1 1 < x 1 < 1 0 < x < 2. We can see that 0 < x < 2 implies the following inequalities: 0 > 3x 2 > 12 and 0 > 3x > 6. Putting these together gives us the following: 18 < 3x 2 3x < 0 20 < 3x 2 3x 2 < 2. Since this inequality gives that 3x 2 3x 2 is negative for x 1 < 1, we must find that 3x 2 3x 2 = ( 3x 2 3x 2), and thus 2 < 3x 2 3x 2 < 20. Thus, we have bounded 3x 2 3x 2. Going back to our previous argument, as long as we choose δ / 3x 2 3x 2, it should be a good choice of δ. Since 3x 2 3x 2 < 20 when x 1 < 1, we might then choose { } 20, 1. 11

For Fun! A big question we might ask is, when considering the limit of a polynomial, will we always be able to use the factoring trick? Much to our delight, the answer is yes! To see this, consider what it means to be able to factor out a (x c) from a polynomial. It means that the polynomial is equal to zero when we plug c in. If we have some polynomial a n x n + a n 1 x n 1 +... + a 1 x + a 0, then consider the new polynomial (a n x n +a n 1 x n 1 +...+a 1 x+a 0 ) (a n c n +a n 1 c n 1 +...+a 1 c+a 0 ), which can be rewritten a n (x n c n ) + a n 1 (x n 1 c n 1 ) +... + a 1 (x c) + a 0 (1 1). This is a polynomial that zeros out when we plug in c, so we should be able to factor out (x c). Suppose we were considering the limit of the original polynomial as x approaches c, then. lim (a nx n + a n 1 x n 1 +... + a 1 x + a 0 ). x c We would guess that the limit is exactly what happens when we plug c in i.e., lim (a nx n +a n 1 x n 1 +...+a 1 x+a 0 ) = (a n c n +a n 1 c n 1 +...+a 1 c+a 0 ). x c When we try to do an epsilon-delta proof to show this is true, given > 0 we would be considering (a n x n +a n 1 x n 1 +...+a 1 x+a 0 ) (a n c n +a n 1 c n 1 +...+a 1 c+a 0 ) <, which can be rewritten a n (x n c n ) + a n 1 (x n 1 c n 1 ) +... + a 1 (x c) + a 0 (1 1). This is precisely the polynomial we argued we could factor (x c) out of, though! This means that whenever we re considering x c < δ for f(x) a polynomial, we can factor a x c out of the f(x) L. (Swank.) 12

5. Prove that lim at x = 2.) 4x+1 x 2 3x 4 4x+1 = 9/2. (This proves that 3x 4 is continuous Proof: Let > 0 be given. We wish to find some δ > 0 such that for any x R, 0 < x 2 < δ forces (4x + 1)/(3x 4) 9/2 <. We claim that { 2 19, 1 } 3 would work. To check this, suppose x 2 < { 2 19, 1 3}. Simplifying (4x + 1)/(3x 4) 9/2, we get 4x + 1 3x 4 9 2 = 2(4x + 1) 9(3x 4) (3x 4)2 = 19(x 2) 2(3x 4) Using our choice of δ, we then notice 19(x 2) 2(3x 4) = x 2 19 2(3x 4) < x 2 19 ( 19 2 2(3x 4) < 19 2 if x 2 < 1 ) 3 ( ) ( ) ( 2 19 < x 2 < 2 ) 19 2 19 =. Thus, this δ forces (4x+1)/(3x 4) 9/2 < whenever 0 < x 2 < δ. 4x+1 It follows that lim x 2 3x 4 = 9/2. 13

Discussion How would we find δ this time? We have the two inequalities x 2 < δ and (4x + 1)/(3x 4) 9/2 <. Let s see if we can massage our inequality into something that looks more helpful. We did that in our proof above: 4x + 1 3x 4 9 2 = x 2 19 2(3x 4) We note now that our inequalities each have a common coefficient: x 2 < δ and x 2 19 2(3x 4) <. This common coefficient is a good in for understanding how one of our inequalities might affect the other. We note that if we were to choose δ = 2(3x 4) 19, then we would certainly have x 2 < δ = 2(3x 4) 19 x 2 19 2(3x 4) <. However, something we are getting well used to seeing is that this choice of δ depends on x. Thus, we need another δ. Since any time we choose a value for δ that works a smaller one will also work, we can request δ 1, as we have done time and time again before. This gives us x 2 < 1 1 < x 2 < 1 1 < x < 3. We then see that 1 < x < 3 implies the following: 1 < x < 3 3 < 3x < 9 1 < 3x 4 < 5. If we were to act as we have in the previous examples, we would then want to say something to the effect of since 1 < 3x 4, we can choose δ = 2( 1)/19 < 2(3x 4)/19. However, is this actually true for all 1 < x < 3? Unfortunately, no. If we plug in x = 4/3, we find that our δ would have to be less than zero, which is no good. So it is the value x = 4/3 that is screwing us over because our upper bound for δ, 2(3x 4)/19, zeros out there. 14

The question, then, is what can be done about it. Well, we declared δ 1 so that we would have x 2 < 1. The problem was that we found 4/3 2 = 2/3 < 1. We then might want to make a tighter bound on δ such that 4/3 2 is not less than δ. Since 4/3 2 = 2/3, we might then choose δ 1/3. So now we go about the steps we took for δ 1, except now use δ 1/3. Since δ 1/3 and x 2 < δ, then x 2 < 1/3. We then unravel this absolute value inequality: x 2 < 1 3 1 3 < x 2 < 1 3 5 3 < x < 7 3. We then see that 5 3 < x < 7 3 implies the following: 5 3 < x < 7 3 5 < 3x < 7 1 < 3x 4 < 3. Now we have that 1 < 3x 4 < 3, which implies that 3x 4 is positive. This is fantastic, because we can then write 3x 4 = 3x 4, and thus 1 < 3x 4 < 3 Since we have that 1 < 3x 4, we might then choose δ = 2(1) 19 < 2(3x 4)/19. This is the same δ as before, it would appear. However, note that this time we have required that δ 1/3 rather than the δ 1 of before. Thus, the δ we have chosen the second time is { 2 19, 1 } 3 instead of {2/19, 1}. While they are subtly different, we can see that the difference does matter one value of δ may still work while we saw that the other would not! It would then remain to check that our chosen value of δ gets the job done, as we did in the proof above. 15