The Epsilon-Delta Limit Definition:A Few Examples

Nick Rauh

1. Prove that limx→a

x2 = a2. (Since we leave a arbitrary, this is the

same as showing x2 is continuous.)

Proof: Let ε > 0. We wish to find δ > 0 such that for any x ∈ R,0 < |x− a| < δ implies |x2 − a2| < ε. We claim that the choice

δ = min{

ε

|2a|+ 1, 1

}is an appropriate choice of δ. First note that |2a| is always nonneg-ative, so |2a| + 1 is always positive. This means ε

|2a|+1 > 0 for anya ∈ R, so we have actually chosen δ > 0. Next we observe that if|x− a| < δ = min

{ε

|2a|+1 , 1}

, then

|x2 − a2| = |x + a||x− a| (factoring)< (|2a|+ 1)|x− a| (|x + a| < |2a|+ 1 if |x− a| < 1)

< (|2a|+ 1)ε

|2a|+ 1

(|x− a| < ε

|2a|+ 1

)= ε.

Thus, we see that this choice of δ forces |x2 − a2| < ε whenever0 < |x− a| < δ. It follows that lim

x→ax2 = a2. �

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Discussion Once we have a δ > 0 that we think could work, all we have to do isplug it in as we did above and verify. But how did we come up withour particular choice of δ > 0? We basically had to work backwards.We started with the idea that given ε > 0 we want δ > 0 such thatfor any x ∈ R, 0 < |x− a| < δ implies |x2 − a2| < ε. We then had twoinequalities to work with and somehow wanted to use them to comeup with an expression for δ.

We might first note that we can factor |x2−a2| into |x+a||x−a|. Wecould then rewrite our task to be finding a δ0 such that |x − a| < δimplies |x + a||x− a| < ε. The nice thing about this new task is thatwe have a common coefficient in both of our inequalities:

|x− a| < δ ⇒ |x + a| |x− a| < ε

Perhaps this could be our “in” to combining the inequalities and get-ting an expression for δ. Our first step towards this would be isolatingthe common coefficient in our expression for ε:

|x− a| < ε

|x + a|.

We then note that if we were to choose δ ≤ ε/|x + a|, then it wouldcertainly be the case that

|x− a| < δ ≤ ε

|x + a|⇒ |x− a||x + a| < ε.

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 achoice of δ > 0 such that δ does not depend on x.

We know that choosing δ ≤ ε/|x + a| will get the job done. The ques-tion is how we can choose δ in this way. Well, we know that wheneverwe have one δ > 0 that works, any smaller value will also work. Thismeans 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, thisis the same as saying that “x is no further away from a than a dis-tance of 1.” If x is no further away from a than a distance of 1, what

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can we say about |x + a|? Well, we can say that x + a is no furtheraway 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|, whichwas the term in δ = ε/|x + a| that made this choice of δ dependent onx. We note that if we replace the denominator with |2a|+ 1, we get

|x− a| < δ = min{

ε

|2a|+ 1, 1

}<

ε

|x + a|.

The reason that we have to take the minimum of ε/(|2a| + 1) and 1is 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 othervalue—say, δ < 20—then we would find that |x+a| < |2a|+20 on thecondition that |x− a| < δ ≤ 20, and thus choose

δ = min{

ε

|2a|+ 20, 20

}.

We have thus constructed a δ > 0 that we think will work. All thatis left for us to do is go back and make sure that for this δ > 0,0 < |x− a| < δ implies |x2 − a2| < ε, as we did in the proof.

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2. Prove that limx→2

x3 = 8. (This proves that x3 is continuous at

x = 2.)

Proof: Let ε > 0. We want δ > 0 such that for any x ∈ R,0 < |x− 2| < δ forces |x3 − 8| < ε. We claim that

δ = min{ ε

19, 1

}gets the job done. To see that this, suppose |x− 2| < δ = min

{ε19 , 1

}.

Then we must have

|x3 − 8| = |x− 2||x2 + 2x + 4| (factoring)< |x− 2|19 (|x2 + 2x + 4| < 19 if |x− 2| < 1)

<ε

1919

(|x− 2| < ε

19

)= ε.

Thus, we have that this δ forces |x3−8| < ε whenever 0 < |x− 2| < δ.It follows that lim

x→2x3 = 8. �

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Discussion Again, we have created a δ > 0 as if by magic. How would we goabout finding such a δ? As last time, we were given ε > 0 and wantedδ > 0 such that for any x ∈ R, 0 < |x− 2| < δ implies |x3 − 8| < ε.We again had two inequalities to work with and somehow wanted touse them to create an expression for δ.

Since we are again dealing with a polynomial, our gut instinct is toattempt to factor it. This gives |x3−8| = |x−2||x2 +2x+4|, allowingus to rephrase our question. We want δ > 0 such that |x − 2| < δimplies |x − 2||x2 + 2x + 4| < ε. Just like last time we might thenconsider

δ =ε

|x2 + 2x + 4|,

since

|x− 2| < δ =ε

|x2 + 2x + 4|⇒ |x− 2||x2 + 2x + 4| < ε.

However, this choice of δ is dependent on x, which is no good. Wecan save ourselves the same way we did last time, though. If we havea good choice of δ > 0, then any smaller choice will also work, so wecan 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 areequivalent.”) Since |x2 + 2x + 4| is the part of our δ expression thatis causing problems, we will try and use this restriction 1 < x < 3 torestrict |x2 + 2x + 4|.

We begin by noting that 1 < x < 3 implies the following two inequal-ities:

2 < 2x < 6 and 1 < x2 < 9.

Putting these together gives us the following:

3 < x2 + 2x < 15 ⇔ 7 < x2 + 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 thatx is positive. If x is positive, then so is x2 + 2x + 4, so we must have

x2 + 2x + 4 = |x2 + 2x + 4| ⇔ 7 < |x2 + 2x + 4| < 19.

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In this way we have bounded |x2 +2x+4|. Going back to our previousargument, we note that as long as we choose δ ≤ ε/|x2 + 2x + 4|, itshould be a good choice of δ. Since |x2 +2x+4| < 19 when |x−2| < 1,we might then choose

δ = min{ ε

19, 1

},

since 19 > |x2 + 2x + 4| makes

|x− 2| < δ ≤ ε

19<

ε

|x2 + 2x + 4|.

To formulate a proof, it would then remain for us to show that this δgives |x3 − 8| < ε whenever |x− 2| < δ, as we did in the proof above.

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3. Prove that limx→1

|x + 2| = 3. (This shows |x + 2| is continuous at

x = 1.)

Proof: Suppose we are given ε > 0. We then wish to find δ > 0 suchthat for any x ∈ R, 0 < |x− 1| < δ implies ||x + 2| − 3| < ε. We claimthat the choice

δ = min{ε, 1}

is appropriate. To verify this, suppose |x− 1| < δ = min{ε, 1}. Then

|x− 1| < ε ⇒ −ε < x− 1 < ε (unravelling the absolute value)⇒ 3− ε < x + 2 < 3 + ε (adding 3 to all sides)⇒ 3− ε < |x + 2| < 3 + ε (|x + 2| > 0 if |x− 1| < 1)⇒ −ε < |x + 2| − 3 < ε (subtracting 3 from all sides)⇒ ||x + 2| − 3| < ε (combining the inequalities)

Thus, we see that this choice of δ forces ||x + 2| − 3| < ε whenever0 < |x− 1| < δ. Thus, lim

x→1|x + 2| = 3. �

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Discussion Again, we found a δ > 0 that works, so let’s see how we might havefound it. Unlike last time, we have no polynomials to factor. We do,however, have an excessive number of absolute value symbols. When indoubt, unravel those things. We’ll start (as before) with our inequalityfor ε:

||x + 2| − 3| < ε ⇒ −ε < |x + 2| − 3 < ε

⇒ 3− ε < |x + 2| < 3 + ε.

We now have an absolute value wedged between two values. If weunravel this thing, we actually end up with two sets of inequalities.The first type of inequality we have seen before:

−(3 + ε) < x + 2 < 3 + ε ⇔ −5− ε < x < 1 + ε.

The second type is something new:

x + 2 > 3− ε or x + 2 < −(3− ε).

The tricky part in the reasoning is then as follows. We are of themindset that ε is going to be arbitrarily small, so for the values ofε that are worth considering, −(3 − ε) < 0. However, if we set therequirement δ ≤ 1, which is completely within our right, we wouldthen have |x− 1| < 1, so x must be positive. If x > 0, then certainlyx + 2 > 0. Thus, in the above inequality, the option

x + 2 < −(3− ε)

makes no sense, since it claims a positive number is less than a negativenumber. We must then choose

x + 2 > 3− ε ⇔ x > 1− ε.

Thus, we have the following two sets of inequalities:

−5− ε < x < 1 + ε and x > 1− ε.

We can then try to combine them. We note that since −5− ε < 1− εand 1− ε < x, we can combine them in the following way:

−5− ε < 1− ε < x < 1 + ε ⇔ 1− ε < x < 1 + ε

⇔ −ε < x− 1 < ε.

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Since one of our bounds is the negation of the other, we may then foldthis back into an absolute value inequality:

|x− 1| < ε.

It might then seem like ε 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

δ = min{ε, 1}.

To complete a proof, we would then check whether this choice of δ issuch that ||x + 2| − 3| < ε whenever 0 < |x− 1| < δ, as we did in ourproof above.

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4. Prove that limx→1

(4 + x − 3x3) = 2. (This proves that 4 + x − 3x3

is continuous at x = 1.)

Proof: Suppose ε > 0. We want to find some δ > 0 such that for anyx ∈ R, 0 < |x− 1| < δ forces |(4 + x− 3x3)− 2| < ε. We claim that

δ = min{ ε

20, 1

}would be an alright choice for δ. To check this, suppose |x− 2| < δ = min

{ε20 , 1

}.

Then we see that

|(4 + x− 3x3)− 2| = | − 3x3 + x + 2| (simplification)= |x− 1|| − 3x2 + x + 2| (factoring)< |x− 1|20 (| − 3x2 + x + 2| < 20 if |x− 1| < 1)

<ε

2020

(|x− 1| < ε

20

)= ε.

Thus, we have that this δ forces |(4 + x − 3x3) − 2| < ε whenever0 < |x− 1| < δ. Thus, lim

x→1(4 + x− 3x3) = 2. �

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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 − 3x3) − 2| < ε whenever0 < |x− 1| < δ, so we are (as always) attempting to coerce somethinguseful out of two inequalities. Since we are back to dealing with poly-nomials, our gut instinct is to factor. Our polynomial is even messierthan the previous two we have looked at, but we argue that we canstill use a factoring trick. We note that

|(4 + x− 3x3)− 2| = | − 3x3 + x + 2| = |x− 1|| − 3x2 − 3x− 2|.

Now our problem is equivalent to wanting δ > 0 such that |x− 1| < δimplies |x− 1|| − 3x3 + x + 2| < ε. Our natural first guess for δ is

δ =ε

| − 3x2 − 3x− 2|,

since

|x−1| < δ =ε

| − 3x2 − 3x− 2|⇒ |x−1||−3x2−3x−2| < ε.

However, just as with our two previous polynomial examples, this δis dependent on x, so we need a different δ. Again, if we have aδ > 0 that works, then any smaller choice will also work, so place therestriction 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 > −3x2 > −12 and 0 > −3x > −6.

Putting these together gives us the following:

−18 < −3x2 − 3x < 0 ⇔ −20 < −3x2 − 3x− 2 < −2.

Since this inequality gives that −3x2−3x−2 is negative for |x−1| < 1,we must find that | − 3x2 − 3x− 2| = −(−3x2 − 3x− 2), and thus

2 < | − 3x2 − 3x− 2| < 20.

Thus, we have bounded | − 3x2 − 3x− 2|. Going back to our previousargument, as long as we choose δ ≤ ε/| − 3x2 − 3x − 2|, it should bea good choice of δ. Since | − 3x2 − 3x− 2| < 20 when |x− 1| < 1, wemight then choose

δ = min{ ε

20, 1

}.

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For Fun! A big question we might ask is, when considering the limit of a poly-nomial, will we always be able to use the factoring trick? Much toour delight, the answer is yes! To see this, consider what it meansto be able to factor out a (x − c) from a polynomial. It means thatthe polynomial is equal to zero when we plug c in. If we have somepolynomial anxn + an−1x

n−1 + . . . + a1x + a0, then consider the newpolynomial

(anxn +an−1xn−1 + . . .+a1x+a0)− (ancn +an−1c

n−1 + . . .+a1c+a0),

which can be rewritten

an(xn − cn) + an−1(xn−1 − cn−1) + . . . + a1(x− c) + a0(1− 1).

This is a polynomial that zeros out when we plug in c, so we shouldbe able to factor out (x − c). Suppose we were considering the limitof the original polynomial as x approaches c, then.

limx→c

(anxn + an−1xn−1 + . . . + a1x + a0).

We would guess that the limit is exactly what happens when we plugc in—i.e.,

limx→c

(anxn+an−1xn−1+. . .+a1x+a0) = (ancn+an−1c

n−1+. . .+a1c+a0).

When we try to do an epsilon-delta proof to show this is true, givenε > 0 we would be considering

|(anxn+an−1xn−1+. . .+a1x+a0)−(ancn+an−1c

n−1+. . .+a1c+a0)| < ε,

which can be rewritten

|an(xn − cn) + an−1(xn−1 − cn−1) + . . . + a1(x− c) + a0(1− 1)|.

This is precisely the polynomial we argued we could factor (x− c) outof, though! This means that whenever we’re considering |x − c| < δfor f(x) a polynomial, we can factor a |x − c| out of the |f(x) − L|.(Swank.)

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5. Prove that limx→2

4x+13x−4 = 9/2. (This proves that 4x+1

3x−4 is continuous

at x = 2.)

Proof: Let ε > 0 be given. We wish to find some δ > 0 such that forany x ∈ R, 0 < |x− 2| < δ forces |(4x + 1)/(3x − 4) − 9/2| < ε. Weclaim that

δ = min{

2ε

19,13

}would work. To check this, suppose |x− 2| < δ = min

{2ε19 , 1

3

}. Sim-

plifying |(4x + 1)/(3x− 4)− 9/2|, we get∣∣∣∣4x + 13x− 4

− 92

∣∣∣∣ =∣∣∣∣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|∣∣∣∣ −192(3x− 4)

∣∣∣∣< |x− 2|19

2

(∣∣∣∣ −192(3x− 4)

∣∣∣∣ <192

if |x− 2| < 13

)<

(2ε

19

) (192

) (|x− 2| < 2ε

19

)= ε.

Thus, this δ forces |(4x+1)/(3x−4)−9/2| < ε whenever 0 < |x− 2| < δ.It follows that lim

x→2

4x+13x−4 = 9/2. �

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Discussion How would we find δ this time? We have the two inequalities |x−2| < δand |(4x + 1)/(3x − 4) − 9/2| < ε. Let’s see if we can massage our εinequality into something that looks more helpful. We did that in ourproof above: ∣∣∣∣4x + 1

3x− 4− 9

2

∣∣∣∣ = |x− 2|∣∣∣∣ −192(3x− 4)

∣∣∣∣We note now that our inequalities each have a common coefficient:

|x− 2| < δ and |x− 2|∣∣∣∣ −192(3x− 4)

∣∣∣∣ < ε.

This common coefficient is a good “in” for understanding how one ofour inequalities might affect the other. We note that if we were tochoose

δ =∣∣∣∣2(3x− 4)

−19

∣∣∣∣ ε,

then we would certainly have

|x− 2| < δ =∣∣∣∣2(3x− 4)

19

∣∣∣∣ ε ⇒ |x− 2|∣∣∣∣ −192(3x− 4)

∣∣∣∣ < ε.

However, something we are getting well used to seeing is that thischoice of δ depends on x. Thus, we need another δ. Since any time wechoose a value for δ that works a smaller one will also work, we canrequest δ ≤ 1, as we have done time and time again before. This givesus

|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 thenwant to say something to the effect of “since −1 < 3x − 4, we canchoose δ = |2(−1)/19| ε < |2(3x− 4)/19| ε.” However, is this actuallytrue 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 upperbound for δ, |2(3x− 4)/19| ε, zeros out there.

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The question, then, is what can be done about it. Well, we declaredδ ≤ 1 so that we would have |x − 2| < 1. The problem was that wefound |4/3 − 2| = 2/3 < 1. We then might want to make a tighterbound 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 tookfor δ ≤ 1, except now use δ ≤ 1/3.

Since δ ≤ 1/3 and |x − 2| < δ, then |x − 2| < 1/3. We then unravelthis absolute value inequality:

|x− 2| < 13

⇔ −13

< x− 2 <13

⇔ 53

< x <73.

We then see that 53 < x < 7

3 implies the following:

53

< x <73

⇔ 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|, andthus

1 < |3x− 4| < 3

Since we have that 1 < |3x− 4|, we might then choose

δ =∣∣∣∣2(1)

19

∣∣∣∣ ε < |2(3x− 4)/19| ε.

This is the same δ as before, it would appear. However, note that thistime we have required that δ ≤ 1/3 rather than the δ ≤ 1 of before.Thus, the δ we have chosen the second time is

δ = min{

2ε

19,13

}instead of δ = min {2ε/19, 1}. While they are subtly different, we cansee that the difference does matter—one value of δ may still work whilewe saw that the other would not! It would then remain to check thatour chosen value of δ gets the job done, as we did in the proof above.

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