Box 2.2a: Proof of the Sum of Limits

First, recollect what we mean by a limit. Remember that it is like a contract -- a contract to come close to something. I tell you how close to come, and you tell me what I have to do to get that close.

We shall prove that the sum of two limits is equal to the limit of the sum for both sequences and for real valued functions of a real variable. The proof for sequences is first.

Suppose we have two sequences, s_k and t_k, which both have limits, L_s and L_t respectively. Just so that you are not confused by the notation, s_k does not represent a single number, but an infinite sequence of real numbers. The subscript, k ranges through all the counting numbers, starting at 1. So s₁ is the first in the sequence, s₂ is the second, and so on. t_k a represents different sequence in exactly the same way. If you feel intimidated by the notation, please stop and reread this paragraph until you feel more confident.

Recalling the contract analogy, what contracts do the limits, L_s and L_t impose upon the sequences s_k and t_k? See if you can write the essence of these contracts on a piece of paper before continuing. Then click here to continue.

The contract for s_k is that if I tell you how close s_k has to be to L_s (and remember how we specify closeness? We say that for some positive value of closeness, e_s, it is true that |s_k - L_s| £ e_s) then you can tell me how big k has to be so that s_k and all terms in the sequence that follow it are all at least that close. The contract for t_k is virtually the same -- just substitute t's in for s's. And remember that these contracts must hold for any positive value of e I might name, no matter how small.

What a proof of the sum rule for limits must show is that given that the above contracts hold, that a similar contract for the sum holds. What does that contract look like? See if you can write the essence of that one down on a piece of paper before continuing. Then click here to continue.

In case you had trouble with finding the contract for the sum, here it is: if I give you any degree of closeness, e_plus, that s_k + t_k is required to be to L_s + L_t, then you can tell me how big k has to be so that it and all terms that follow in the sequences do indeed come that close. In math symbols this translates to: given any positive e_plus, no matter how small, there exists an n such that:

   |(s_k + t_k) - (L_s + L_t)| £ e_plus

whenever k ³ n. (Please make sure you understand how the symbolic statement says the same thing as the statement above it using words. See if you can write similar symbolic statements for the first two contracts)

So if we can show how the first two contracts lead inescapably to the third, then we have proved that the sum of the limits is equal to the limit of the sum.

I'd like you to take a stab at finishing the proof yourself. Don't be discouraged if you get stuck, but please do make a serious effort, and don't give up at the first sign of rough seas. If you do get stuck, try taking a 5 minute break and then try again. When you are done, or when you are certain that your efforts have reached the point of dimishing returns, click here to see the step-by-step of the proof. Your hint is this inequality, which holds for all real numbers, a and b:

     |a + b|  £  |a| + |b|

and the fact that if a £ b and b £ c, then a £ c. This, again, holds for all real numbers, a, b, and c, and it is called the transitivity of less than or equal to. (can you name some other relationships that are transitive?)

Step 1) Observe that the two original contracts give us two inequalities:

   |s_k - L_s|  £  e_s

whenever k ³ n_s, and

   |t_k - L_t|  £  e_t

whenever k ³ n_t. I can hear you saying, "What's this n_s and n_t stuff? What part of his body did Karl pull those names out of?" Well, remember that the contracts say that if I give you a degree of closeness I need to see, then you can tell me how far into the sequence to go to be that close. That "how far" may not be the same for both sequences. So n_s is the "how far" for the s_k sequence and n_t is the "how far" for the t_k sequence.

Step 2) Write the inequality we must derive from the above (I already gave you this one a little while ago, but here it is again): given any positive e_plus, no matter how small, there exists n such that

   |(s_k + t_k) - (L_s + L_t)| £ e_plus

whenever k ³ n.

Step 3) Establish our e's. What I mean by this is that for each contract, the e is arbitrary. I tell you what I how small I want it to be, and you tell me what I have to do to get it that small. So if I tell you that I want e_plus to be such and such, you can tell me, make both e_s and e_t equal to half of that in the first two contracts. I say, "Ok, let's do that, so how deep into the sequences do I have to go to make that happen?" You tell me, n_s for the s_k sequence and n_t for the t_k sequence. So we have:

                 e_plus
   |s_k - L_s|  £       
                   2

whenever k ³ n_s, and

                 e_plus
   |t_k - L_t|  £       
                   2

whenever k ³ n_t. We know that n_s and n_t must exist because we know (by assumption) that the first two contracts hold.

Step 4) Pick the larger of the two n's. I have to pick just one of those n's (you will see why shortly. The proof relies on making both contracts hold under identical conditions). Which one guarantees that both contracts will hold? Well remember the way the contracts are worded: "whenever k is greater than or equal to n_whatever." If k is greater than or equal to the larger of the two n's, then it is certainly greater than or equal to both of them. So pick the larger of the two and call it n. Based upon that, we now have:

                 e_plus
   |s_k - L_s|  £       
                   2

and

                 e_plus
   |t_k - L_t|  £       
                   2

whenever k ³ n.

Step 5) Add the two inequalities. We can do that because they both hold under identical conditions now. So we have:

   |s_k - L_s|  +  |t_k - L_t|  £  e_plus

Step 6) Apply the hints. Remember the ones I gave you before I asked you to go off and try this proof by yourself? We can see that on the right hand side of the last inequality we have the sum of two absolute value expression. The first part of the hint tells us that:

   |(s_k - L_s) + (t_k - L_t)|  £  |s_k - L_s|  +  |t_k - L_t|

and this holds unconditionally.

Now apply transitivity of less than or equal to to the two inequalities you have here (if you forgot what transitivity is, go back and look at it again). From transitivity we get:

   |(s_k + t_k) - (L_s + L_t)| £ e_plus

whenever k >= n, and we have shown how to find n using the first two contracts.

Where have we seen this bad boy before? Remember -- it's our third contract -- what we have been trying to show is an inescapable consequence of the first two contracts. And that means that the proof is complete, because we have logically deduced the third contract from the first two. So if I tell you, "Make the sum within e_plus of the sum of the limits, L_s and L_t," you can pick the two e's for the first two contracts to be half of that, establish the two n's, pick the larger of them, and tell me that that is how far into the sequences I have to go.

Of course, if you have to produce this proof on an exam, you won't have to be nearly as wordy as I have been. You will have to be able to reproduce the statements in math symbols, and you will do well by providing a sentence or two explaining the how and why of each step. And to do that, you must understand the proof. So review it till you do.

Now we are going to move on to the proving that for real valued functions of a real variable, the sum of the limits is equal to the limit of the sum. As I have already stated, this type of limit is very similar (although it may not seem so at first glance) to the limit of a series. So, not surprisingly, the proof is very similar to the proof you have just waded through. So I will dispense with a lot of the word-explanations I gave in the last proof and do more of a pure math proof. If you understood the last one, you should be able to follow along on this one. If you have trouble with this one, go back and review the last one again.

In keeping with the contract analogy, with this type of limit we have similar contracts, but with slightly different wording.

Recall how we defined:

     lim  f(x)  =  L_f
    x  > a

The contract said, if I tell you how close f(x) has to be to L_f, then you can tell me how close x has to be to a in order to make that happen. We shall express the closeness of f(x) to L_f using e_f and the closeness of x to a using d_f. And we have, in math lingo:

For every e_f > 0 we can always find d_f > 0 such that |f(x) - L_f| £ e_f whenever |x - a| £ d_f

Now suppose we have a second function, g(x) that has a limit as x goes toward a:

     lim  g(x)  =  L_g
    x  > a

So, take everything we just stated and replace the f's with g's, and you have the equivalent for this second function:

For every e_g > 0 we can always find d_g > 0 such that |g(x) - L_g| £ e_g whenever |x - a| £ d_g

We aim to prove that an inescapable consequence of these two statements is

For every e_plus > 0 we can always find d_plus > 0 such that |f(x) + g(x) - (L_f + L_g)| £ e_plus whenever |x - a| £ d_plus.

At this point, I'd like to invite you to complete the proof yourself. If you run into problems, go back and review the last proof. You should know that the proof that follows is more likely to be on the exam than the one we have already done. You need to be prepared to produce it. So try hard. When you are done, see if what you have is close to what I have by clicking here. The same hints apply here as in the last proof.

Step 1) Choosing our e's. Remember that the values for e_f and e_g are arbitrary. I tell you what one of them is, and you tell me what the corresponding d is -- that's the contract. So, if for the sum, I give you a value of e_plus, you can say, make both e_f and e_g equal to half of that. Now our contracts look like:

We can find d_f > 0 such that |f(x) - L_f| £ e_plus / 2 whenever |x - a| £ d_f

Likewise with g(x) we have:

We can find d_g > 0 such that |f(x) - L_g| £ e_plus / 2 whenever |x - a| £ d_g

Step 2) choosing our d. Again we have to make both inequalities hold under identical conditions. So we take the lesser of d_f and d_g and we call that d_plus. And from that we get:

We can find d > 0 such that |f(x) - L_f| £ e_plus / 2 and |f(x) - L_g| £ e_plus / 2 whenever |x - a| £ d_plus

Why are we justified in making this substitution? Because if |x - a| is less than or equal to the lesser of the two d's, then it is certainly less than or equal to both of them.

Step 3) Add the inequalities. Now we have two inequalities that hold under identical conditions. That means we can add them.

   |f(x) - L_f|  +  |g(x) - L_g|  £  e_plus

Step 4) Apply the hints. The inequality in the hint allows us that

   |(f(x) - L_f) + (g(x) - L_g)|  £  |f(x) - L_f|  +  |g(x) - L_g|

Finally, applying transitivity we have:

|(f(x) - L_f) + (g(x) - L_g)|  £  e_plus

whenever |x - a| £ d_plus, which is exactly the contract we want to show. And we have a perscription for finding d_plus. That means that this last contract is an inevitable consequence of the first two, and that means the theorem is proved.

I can't overstate how important it is that you are comfortable with the line of thinking that goes into the two proofs you have just been through. It may be for you that this is still be like trying to bench press that 200 lb. barbell only to have it descend slowly to your chest and stay there. If that's where you find yourself, just keep going over and over the material in this box. Also go back and reread the story about the boat in the fog. Reread the material in 2.0 and 2.1 that introduces you to limits. Keep at it. You can lift that barbell, and you can understand this stuff, but only if you keep working at it. You are not the first to have difficulty with this, and when you understand it, you will not be the first to overcome your difficulties. Countless multitudes have been here before, and countless multitudes of folks just like you have overcome.

As always, I am at hahn@netsrq.com

Return to main text