Now in order to do so, there's one difficult and critical step.

That is, you need to determine the recursion relation that the terms satisfy.

In this case, I'll tell you that it is the following.

a sub n = square root of 2 + a sub n-1.

Let's say I give you that recursion relation.

How do you compute the limit?

Let us, as before, denote this limit by L and

then apply the limit to the recursion relation.

On the left, we have the limit, as n goes to infinity, of a sub n.

That's L.

On the right we have an a sub n-1 term.

What's the limit of that as n goes to infinity?

Well of course, it too must be L.

Now we're assuming that we can slip that limit in under the square root,

let's forget about whether that's legal or not, and

see what happens when we try to simplify this algebraically.

Squaring both sides, we obtain L squared = 2 + L.

With a little bit of rearrangement, we get a polynomial

that easily factors into (L-2) and (L+1).

Now, this polynomial has two solutions,

namely negative 1 and positive 2.

We know that the limit of this sequence is not a negative number.

And therefore, if the limit exists, it must be equal to 2.

Indeed the limit does exist and is 2.

Let's look at another limit of this form.

In this case, 1+1 over 1+1 over 1+1,

etc., all nested together.

As before we can write out a sequence that gets closer and closer to this.

a naught is 1, a1 is 1 over 1 + 1, a2 is 1 over 1 +

1 over 1 over 1 or something like that, I don't know.

It keeps going.

The tricky part in this case is to find

the recurrence relation that these terms satisfy.

Once again, I’m going to tell you what it is.

It is that a sub n = 1 + 1 over a sub n-1.

This instruction tells you how to build the next term.

In order to compute the limit of the a sub ns, we follow the same

procedure as before, taking the limit of the recursion relation,

substituting in L for a sub n, and a sub n-1.

And then, with a little bit of algebraic rearrangement, what do we see?

We get L squared- L- 1 = 0.

This has roots 1 plus or minus square root of 5 over 2.

One of these is negative, the other positive.

We'll take that positive square root, and that is our limit.

This particular value is of some independent interest.

It is sometimes called the golden ratio and given the symbol phi.

It equals 1 + root 5 over 2.

Now perhaps you've seen the golden ratio before in

the context of some other sequences.

Perhaps you've seen the Fibonacci sequence,

that begins with 0, 1, 1, 2, 3, 5, 8,

13, 21, 34, 55, 89, etc.

There's a lot of interesting mathematics hiding behind here.

And if you'd like, you might want to take a look at the bonus material for

a hint at what calculus will be able to tell us about the Fibonacci sequence.