0:20

And one way to think about this is via Taylor series.

And what do Taylor series tell us?

Well Taylor series tells us that sine of x, its equal to x minus x cubed over

6 plus x to 5th over 5 factorial minus, and the series

would keep on going. But at least I know that

sine of x is approximately x plus,

I'll just write higher order terms. So

if you believe this, you might try to evaluate this limit by making use

of this, admittedly at this point, very vague fact.

So what's the limit then as x approaches 0

of x plus some higher order terms divided by x?

Well that's the sort of limit that I could, you know, really approach.

Right?

I mean it's not a transitional function anymore, it's just it

looks like a polynomial as far as I'm concerned or imagining here.

So I could think about how do I calculate

the limit as x approaches 0 of x plus higher order terms over x.

Well what I would probably want to do is multiply

the numerator and the denominator by 1 over x.

And I've got these higher order terms and the denominator's got x.

So this looks like the limit, as x approaches 0

of the numerator now is 1 plus these higher order terms.

And the denominator is just 1.

And if

I'm taking the limit then, as x approaches 0, be

very tempted to say that the limit is just 1.

And that is not even close to a proof.

Well, what's wrong with this argument?

Well, basically, this is a circular argument.

I mean, how so?

So I'm trying to, trying to calculate the limit of sine

x over x by thinking about a Taylor series for sine.

And to find a Taylor series for sine, right, to find

this Taylor series, I need to be able to differentiate sine.

So what do you have to do to be able to differentiate sine?

Well, if you're trying to differentiate sine, you in

particular need to be able to differentiate sine at 0.

So let me just write down the limit of the

difference quotient that calculates the derivative of sine at 0.

That's the limit as h approaches 0 of sine of 0 plus h minus sine of 0 divided by h.

Right,

this limit is calculating the derivative of sine at 0, cause its the ratio

of how the output changes when you go from 0 to 0 plus h.

That's how the input changes when you go from 0 to h.

2:51

Okay. But now, how do I calculate this limit?

Well, what is this limit?

This is this limit, as h goes to 0, of what sine of h minus

sine of 0 oh, sine of 0's 0, so the whole numerator's just sine of h, divided by h.

[LAUGH]

So that's exactly the limit that we've got here, just with x replaced by h.

So what happened here, right?

I'm imagining that I'm trying to calculate

this limit by thinking about Taylor series.

But to calculate the Taylor series for sine,

you have to be able to differentiate sine.

But to be able to differentiate sine, I

really need to know, or calculate, this limit.

So, you know, if you think you're really

proving anything by this method, you're really not.

I mean, this is just a circular argument.

Admittedly, it's a circular argument, in a really big circle!

So maybe it looks like a straight line, or a reasonable

argument, but there really is something essentially circular going on here.

But the fact that the argument's circular, shouldn't stop us

from making use of that kind of thinking where it's appropriate.

Let's find the limit as x approaches 0 of cosine of x minus 1 over the numerator,

divided by sine of x times log of 1 minus x.

That's a very complicated looking limit question.

Let me try to approach that more

complicated limit question by using Taylor series.

So let's think about cosine.

So the Taylor series for cosine looks like this.

Cosine of x is 1 minus x squared over 2 plus, and those higher order terms.

So I'm going to write big O, x to the 4th. And to be a little

more pedantic, I'll say as x approaches 0. And what does this mean?

[LAUGH]

Well, I don't want to talk about this too precisely yet, but

morally, or intuitively at least, you should think about it like this.

Cosine of x is 1 minus x squared over 2 plus, you can just think of

these as being the higher order terms in

the Taylor series expansion of cosine around 0.

Now the next term is an x to the 4th term

but there's also an x to the 6th term and so on.

So what about cosine x minus 1? So that means that cosine

of x minus 1 is negative x squared over 2 plus

more terms of degree at least 4. We can write sine of x in the same way.

Sine of x, now sine also has a Taylor series expansion, it's x.

And then there's higher order terms, right?

So I'm just going to write plus higher order terms starting with an x

cubed term. Let's do the log term.

Well, log of 1 minus x, as a Taylor

series expansion that starts like this, minus x minus x

squared over 2 minus x cubed over 3 minus x

to the 4th over 4 and that keeps on going.

And I could write that as just minus x plus terms of degree at least 2.

Let's put the sine and

log term together.

So let's see, I want to multiply together things that involve these big O's.

So how does that work?

Well I'm trying to calculate sine of x times log of 1 minus x.

I can write down a Taylor series for that, but

I don't want to actually go through the bother of calculating

out what the Taylor series is going to be, so I'm

just going try to multiply together these these, these Taylor series.

So sine of x is x plus terms of degree at least

3, and log of 1 minus x, I mean not everywhere, but at least around 0, is

given by negative x plus terms of degree at least 2.

So now what happens when I multiply together these two Taylor series?

Well I've got an x times a negative x, so that gives me negative x squared.

But what

other terms do I get?

Well, here I've got terms of degree at least 3 times x.

These'll give me terms of degree at least 4.

Things of degree of at least 3 times things of degree

at least 2 will give me thing of degree at least 5.

And I've got x times things of degree of at least 2, so this x

times possibly some x squared terms, that'll give

me somethings of degree of at least 3.

So all told what I know, is that this

Taylor Series, when I multiply them together,

is going to start out minus x squared.

And there's going to be some more terms of degree at least 3.

I'm not bothering to calculate those, right?

They're all sort of hidden in this big O notation.

Now we're in a position to consider the original limit.

So thinking about Taylor series for the numerator and denominator, I might

be tempted or I, this is the limit as x approaches 0.

I've got a Taylor series for cosine of x

minus 1.

It's negative x squared over 2 plus terms, degree at least 4.

And I just found a Taylor series for the denominator by

multiplying together the Taylor series for sine and the Taylor series

for log of 1 minus x, and that ended up being

minus x squared plus some terms of degree at least 3.

Now I could multiply this by just the disguised version

of 1, so 1 over x squared divided by 1 over x squared, and this'll give me what?

The limit as x approaches 0 of this numerator is now minus one

half plus terms of degree at least 2. And the denominator

is now just minus 1 plus terms of degree at least 1.

But what is this limit, right? I'm imagining

that x is approaching 0, so these terms and these terms are both approaching 0,

so the numerator has limit minus one half and the denominator has limit minus 1.

And that means that this limit, the limit of the ratio is just one half.

So maybe, at this point, you're unimpressed.

You're thinking, I could have done that with L'Hopital's rule.

And you'd be right.

I mean, you could have used L'Hopital's rule.

So why are we thinking about Taylor series?

Well, the point here isn't

just that Taylor series are useful computationally.

My claim is that Taylor series is really

providing some deeper insight into what's going on.

So we're presented with this fact, that this limit is equal to one

half and you're probably thinking, who cares, and you'd be right to think that.

Who cares if this limit is equal to one half?

If a machine told me this is equal to one half, I wouldn't care either.

What matters isn't that the machines are telling us that this

is equal to one half or that we've got some horribly

long algebraic calculation that proves to us that this limit is equal to one half.

What really matters here is that this limit is equal

to one half for a reason that human beings can understand.

We can really comprehend why this limit should be equal to one half.

We can think about Taylor series.

And Taylor series are telling us that this numerator looks

like negative X squared over 2 plus higher order terms.

We think about Taylor series for sine, and sine looks

like x plus x cubed over 6 and so on. X plus higher order terms.

The Taylor series for log of 1 minus x,

starts of negative x minus, and then there's more terms.

And when I can multiply together Taylor series, right?

And I multiply together these two series, I'm

getting a Taylor Series for the whole denominator.

And the denominator then looks like

negative x squared plus higher order terms.

And now, I can bring to bear all of the intuition

that I have about limits of rational functions,

limits of polynomials over polynomials, and as x approaches

0, it makes perfect sense then, that if

this is a polynomial, it's really a power series.

But if I'm thinking that this is a polynomial that starts

off negative x squared over 2 plus higher order terms, and

the denominator looks like a polynomial, really a power series, but

it's like a polynomial, negative X squared plus higher order terms.

Well then what happens, right?

The limit is exactly the ratio of these leading terms.

It's negative one half over negative 1.

So the fact that this limit equals one half, you

can understand a reason for it by thinking about Taylor series.

Yeah, wow and that's always the point, right?

The point isn't getting answers, it's getting insight.

The point of mathematics, you know, it isn't truth, it's proof.

Right?

It's not so much whether something's true, but why

is it true.

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