Contents of this Calculus 2 episode:

Sandwich theorem, Squeeze theorem, Estimation, Lower estimate, Upper estimate, Estimation of n-th radical, Estimation of fractions.

Sequence is “stronger” than sequence , if .

We write it like this:

You can visualize it as the stronger sequence tends to infinity faster.

For example, for large n values, n2 looks like this...

and n3 looks like this.

But when 2n, which is much stronger than both of them, enters the scene, ...

well, that significantly diminishes the self-esteem of the previous two sequences.

But 2n is not all-powerful, because 3n is much stronger.

It seems there is plenty of power-struggle in the world of sequences.

You’d better know everybody’s strength.

Now let's get down to business.

Here is a theorem that we will use for finding the limits of more complicated sequences.

The theorem says that if

and and there is , such that for all it holds that

, then .

This theorem is called the Squeeze Theorem (or Sandwich Theorem), and one of its typical application area is finding

type limits.

Let's see this limit, for instance:

We have to realize that the strongest term here is , and

this means that for large n values the other terms are relatively tiny,

as if they weren’t there at all.

So, it shouldn’t be surprising that the limit

The Squeeze Theorem will help us to prove this in a precise manner.

Here is the plan:

First, we find a sequence that is less than the original sequence,

and it has a limit of 5.

Next, we find another sequence that is greater than the original sequence, and also tends to 5.

Finally we conclude with content that the original sequence also tends to 5.

Well, this is a nice plan, but it isn’t that easy to carry out.

It is OK that we need a sequence that is less than the original sequence,

and another one that is greater than the original sequence.

Anybody could come up with sequences like that. However, finding sequences that also tend to 5, well, that is a lot harder.

To do that, we need to learn the art of estimation.

We have to be careful not to modify the strongest term during the estimation process.

When finding the lower estimate, we omit everything but the strongest term.

When finding the upper estimate, we replace all terms with the strongest term.

Let's see another limit.

The strongest term in the numerator is , and in the denominator it is .

And now we are ready for the estimation.

To find the lower estimate, we decrease the numerator, and increase the denominator.

But we have to take care to not modify the strongest term neither in the numerator nor in the denominator.

We decrease the numerator by keeping only the strongest term.

We increase the denominator by replacing all terms with the strongest one.

When finding the upper estimate, we increase the numerator by replacing all terms with the strongest term.

We then decrease the denominator by keeping only the strongest term.

But there are more complicated cases, too:

The upper estimation is fairly easy, we just omit the negative term:

The lower estimation however, is quite cunning.

Here we need something that is less than the original.

So, it is not good to omit 5n, we should actually do the opposite: subtract more than that.

Yes, but what should we subtract? If we subtract - let’s say - 6n, then the lower estimate is zero.

And sadly, that is a problem...

We need to use a trick here.

Well, the lower estimate will be this. Here, C is some positive number, less than 1.

The funny thing is that this is the only requirement.

Let’s say it is 1/2.

Let’s check whether it is less than the original:

Well, yes, if .

From here it is routine work.

Now, what’s next?

Now let’s continue with a few funny limits.

The strongest term is 6n, the others are irrelevant for the limit.

Since , the limit of the sequence is 6.

But this is only an intuition, an idea. For the precise proof, we need a theorem called the Squeeze Theorem.

And now we are ready for the precise solution.

Let’s start with the upper estimate, which happens to be very easy this time.

The lower estimate is more interesting.

We are going to do some hocus-pocus here. First, we replace each negative term with the strongest negative term.

And then we estimate it from below.

Here, C is some positive number, less than 1. The funny thing is that this is the only requirement.

Let’s say it is 1/2.

Well, this holds if .

Both the lower and upper estimates tend to 6, so now the limit of the original sequence is officially confirmed as 6.

Well, this is wonderful news, so let’s check out one more.

The upper estimation is simple. We increase the numerator by omitting the negative terms,

and decrease the denominator by omitting this.

Since is definitely not negative.

The lower estimate is more interesting.

We have to increase the denominator, and that is not a problem.

However, we have to decrease the numerator, and we will use the previous trick for that.

Well, this will hold sooner or later. To be exact, for n values greater than 8.

And now come a few very funny limits.

All of them will be like these:

On top of it, we will also have to use the Squeeze Theorem.

Let’s start with an easy one:

For the upper estimate, we decrease the denominator,

and for the lower estimate, we increase it:

Here comes a little trick.

Now these are the really funny cases.

We would be much happier without this n. As it turns out, that could actually be managed:

And Voila! From here it is exactly the same as the previous problem.

Calculus 2 episode