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How large is large enough?

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[SOUND]

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The definition of limit, says that to get
within the

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epsilon of L, I just have to go past the
big Nth term.

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To guarantee that you're within epsilon,
how big does N need to be?

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We can actually compute this, in some
cases.

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Consider the sequence

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a sub n equals n plus 1 divided by n plus
2.

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So what's the limit?

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The limit of this sequence as n approaches
infinity is 1.

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Let me draw a picture of this.

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So here' s a number line.

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let's put 0 all the way over here, let's
put 1 right here.

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And I've got a whole bunch of terms on the
sequence.

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Right?
So here's the first term.

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Here's the second term.
Here's the third term, and so on.

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And as I go out further and further in the
sequence, the terms get closer

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and closer to 1.
And the question is, how far do I have to

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go out in the sequence, to guarantee that
I'm within some epsilon of 1.

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Now let's suppose that I want to be within

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a 100th of 1.
How big does N have to be?

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So what I want to do, is find a value for
big N.

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So that whenever little n is bigger than
or equal to big N, I get

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that the nth term of my sequence is within
100th of my limit 1.

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Right?

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This is telling me that the nth term is
within epsilon.

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Epsilon being 1 100th in this case, of my
limiting value 1.

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But I can rewrite this.

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Instead of writing it this way, I could
instead write that

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a sub n should be between 99 100ths, and a
101ths.

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to be between 99 and a 101ths, is exactly

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the same thing as being within a 100th of
1.

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Now I've got a formula for a sub n.

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So I could instead

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write, this is 99 over 100, the formula
for a sub n, is n plus

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1 over n plus 2, is less than 101ths.
So, what I'm trying to

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do, is figure out who big I need big N to
be, so that whenever little n

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is bigger than big N, I know that both of
these inequalities hold.

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Meaning that my nth term, is really within

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a 100th of 1.

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Well, one of these inequalities come for
free.

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This inequality here comes for free,
because n plus

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1 over n plus 2, is always less than 1.

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Right?

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The numerator here is smaller than the
denominator.

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So this thing being less than 1, in

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particular, this thing is less than 101
over 100.

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So I get this inequality for free.

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This inequality, however, requires a
little bit of work.

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I could solve here by say multiplying both
sides by n plus 2 and

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by 100, and I end up finding that n needs
to be at least 98.

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So as long as I choose a value for big N,

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which is bigger than 98, that guarantees
that this inequality holds.

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This inequality holds automatically.

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That tells me that my nth term, is really
within

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a 100th of 1.
This is pretty awesome.

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Alright, it's pretty cool that we can tell
if your past, the 98th term in this

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sequence, then you're within a 100th of 1.
And there's nothing special about a 100th.

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If you wanted to be within a billionth of
1,

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you just have to go much further out in
the sequence.

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And then, you get there.

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Right?

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And no matter how close you want to be to
1, if you go far enough out in the

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sequence, you'll be that close.

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And that's exactly what it means, to say
that the limit of this sequence is 1.

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[SOUND]