A Kick in the Discovery

We’ve studied what it means for a sequence to converge to a limit, but we don’t want to be forced to have to prove every limit using the definition! So today, we learn our first tool to prove a limit exists and find it without needing to use the definition of a limit! It’s called the squeeze theorem. I used the squeeze theorem every other day when I took real analysis for the first time. It’s that useful.

Intuition

Imagine we are given some complicated sequence As always, we begin by playing around with the expression defining when we happen to notice that for any the sequence is always larger than and less than It’s somehow sandwiched between and in a manner of speaking.

Motivated by this discovery, let’s define two sequences and

Observe that both and ¹ As these two sequences converge to 1 the and terms are squeezing the towards 1 too since For this reason, we might guess that as well. And we’d be correct! This is the content of the squeeze theorem.

The Squeeze Theorem

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Theorem (The Squeeze Theorem): Let , and be three sequences such that for all If and then as well.

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We will do two proofs, one short and quick. The second longer, but gives good pracitce using the limit definition.

Proof (i): Let , and be three sequences such that for all Assume that both and converge to the same limit, i.e. and We aim to show that as well.

Let By definition, there exists an and such that,

for all and

for all Let Then, for all we have both:

By assumption and therefore,

or better yet for all completing the proof.

Proof (ii): Assume for a contradiction that Then, there exists an such that for all there is some where This means either,

(1) or

(2)

for some Consider case (1). Since we know there exists an such that,

or,

for all But, this contradicts since we’d have

A similar contradiction arises with case (2) using

Infinite Limits

There is a slightly different version of the squeeze theorem. This one is useful to show that a limit diverges to infinity!

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Theorem: (The Infinite Squeeze Theorem): Let and be sequences such that for all If then as well.

Likewise, let and be sequences such that for all If then as well.

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Proof: Let and be sequences such that and for all Consider some Recall in order to prove that we must show that exists an such that for all But, since there exists an such that for all We claim that letting does the job since,

for all

Therefore as well.

For the second situation (when and ) I challenge you to try it out! You got this!!!!!

Closing Remarks

The squeeze theorem is a great utility tool to have under your belt. When we use it we don’t really care about the details of the other than how it compares to two known sequences. This also means the more sequences you know, the more tools you have at your disposal!

Oh, and one final note.

Go back to the section Intuition, where we had two sequences and What if we changed the first 100 terms of so that Say we made for all

Then, since we know for all Sooo… we cannot use the squeeze theorem anymore, right? WRONG! The squeeze theorem still applies to for all since we still have and for all This is a consequence of the fact that limits don’t really care about the beginning of the sequence; limits cares about the end of the sequence.² Changing only the first 100 doesn’t amount to any change in 's limit. So we could have also written the squeeze theorem as:

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Theorem (The Squeeze Theorem (Nuanced)): Let , and be three sequences such that for all where If and then as well.

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Any who, I hope you learned something, had some fun, and/or got a kick in the discovery reading this! If you have any questions, comments, or concerns about the content or my description of it, please leave it in the comments!

Footnote:

1. See part 2 here to learn how to prove these. ↩︎
2. You can prove this statement too! Your goal is to prove the following:
   Let and sequences such that for all If then ↩︎