0:22

For example, here's a problem that you might be

tempted to use l'Hopital on. You know, in a limit as x approaches zero

of sine x over x. Now, what do you have to check?

Well, let's take a look at the limit of the numerator by itself.

The limit of sine x as x approaches zero, sine is continuous so that limit is just

sine of zero, which is zero. Numerator's limit is zero.

Limit of our denominator? the limit of x as x approaches zero is

zero. So, yeah.

This is a zero over zero indeterminate form.

So, we have tried using l'Hopital. l'Hopital tells us that this limit is

equal to the limit as x approaches zero of the derivative of the numerator, which

is cosine x over the derivative of the denominator, which is one.

Now, this limit is quite a bit easier to do, right?

This is really just the limit of cosin x as x approaches zero.

Cosine is continuous so this is just the value of cosine at zero, which is one.

And yeah, I mean, the limit of sine x over x as x

it approaches zero, it is, in fact, one.

This looks good, alright?

The, the question though is how did you know that the derivative of sine X was

cosine x? You needed to know that in order to apply

l'Hopital. How did you know the derivative of sine x

was cosine x? Well, let's think back.

1:46

Right. So, how did I know that the derivative of

sine x was cosine x? Well, taking a derivative is the same

thing as a limit of a difference quotient.

So, to calculate, say, the derivative of sine at zero, I take a limit like this.

The limit if h goes to zero of sine 0 + h - sine 0 / h.

Sine of zero is just zero. So, calculating this limit, which is

calculating the derivative of sine at 0, is really the same as calculating the

limit as h goes to zero, sine 0 + h, that's just sign h - 0 / h.

2:28

Now, what just happened? I wanted to calculate the derivative of

sine at zero, and to calculate the derivative of sine at zero, is really the

same as calculating the limit at h goes to zero of sine h over h.

It's the definition of derivative. And I was going to use this to evaluate

the limit of sine x over x using l'Hopital,

right? l'Hopital doesn't just reduce a limit problem down to another limit

problem. It reduces a limit problem down to a

limit problem and two derivative problems,

right? To, to do l'Hopital, you have to be able

to differentiate the numerator and denominator.

So, if you're going to use l'Hopital to evaluate sine x over x as x goes to zero,

you're going to have to differentiate sine x.

But then, to differentiate sin x at zero, which is what you're trying to do, to, to

use l'Hopital, you have to be able to evaluate this limit, the limit of sine h

over h. You have to be able to do the limit

you're trying to do already. It's just circular reasoning.

So, what just happened? I was trying to calculate the limit of

sine x over x as x approaches zero. I used l'Hopital.

l'Hopita told me to differentiate sine x. But then, to differentiate sine x, I

would've needed to be able to evaluate the limit of sine h over h as h goes to

zero, the exact problem I'm trying to solve.

It's an example of circular reasoning. There's other things that can go wrong

with l'Hopital. One of the conditions of l'Hopital is

that the limit of the ratio of the derivatives has to exist.

let's see an example where that fails to happen.

3:59

Here's an example. Let's take a look at the limit of x plus

sine x over x as x approaches infinity. Try to use l'Hopital.

So, I want to look at the limit of the numerator by itself.

The limit of x plus sine x as x approaches infinity is infinity.

The limit of the denominator by itself, the limit of x as x approaches infinity,

also infinity. This is an infinity over infinity

indeterminate form. It seems good.

Now, l'Hopital's tells me that to calculate this, I should look at the

limit as x approaches infinity. The derivative of the numerator, which is

one plus cosine x over the derivative of the denominator, which is one.

Now, what's the limit of one plus cosine x as x approaches infinity?

Well, the limit of one is just one. But what's the limit of cosine x as x

approaches infinity, right?

Cosine just oscillates between -1 and 1. This limit doesn't exist.

This limit doesn't exist. Now, does that mean that this limit

doesn't exist? I, I, I wrote equals here, but remember,

I'm using l'Hopital's rule, alright.

And what does l'Hopital's rule actually say?

It says, that this limit is equal to the limit of the derivative over the

derivative, provided this limit exists. And this limit doesn't exist in this

case. So, l'Hopital is silent as to the value

of this limit. Let's see if we can actually compute that

limit some other way. Alright.

So, let's try to do this limit some other way.

We're going to do the limit of x plus sine over x as x goes to infinity without

using l'Hopital. This is a limit of fraction, we could

split up the fraction as a limit as x goes to infinity of x over x plus sine x

over x. Now, that's a limit of a sum, which is

the sum of the limits provided the limits exist.

So, this is the limit as x goes to infinity of x over x plus the limit as x

goes to infinity of sine x over x. Now, what's the limit x over x as x goes

to infinity? Just one.

What's the limit of sine x over x as x goes to infinity?

Well, sine x is always between -1 and 1. And x here is going to infinity.

I can make x as big as I like. So, question is can I make this as close

to zero as I like? Yeah.

By making x large enough, a number between -1 and 1 / x can be made as close

to zero as I like. So, this limit is zero.

[SOUND] So, the sum of the limits is one [SOUND] and that's the limit of the sum.

That's the limit of x plus sine x over x. It's one as x goes to infinity.

And this is true in spite of the fact that l'Hopital failed us, right?

When we did l'Hopital's in this problem, I was told to differentiate the

numerator, divided it by the derivative of the denominator as x goes to infinity,

and that derivative didn't exist. It was oscillating between zero and two.

And in that case, l'Hopital is just silent.

l'Hopital requires that limit of the ratio of the derivatives to exist, and if

it doesn't exist, l'Hopital doesn't say anything.

This limit does exist and then to see that requires some, some algebraic

computation. So, we manage to evaluate that limit.

The limit was one, but no thanks to l'Hopital.

The limit of the ratio of the derivatives didn't exist,

it was oscillating. We evaluated that limit by algebraic

manipulation. And don't get me wrong.

l'Hopital is awfully powerful. There's plenty of situations where you

want to use l'Hopital. But often, you can get away with just doing some algebraic

manipulation. So, I encourage you, when you're doing

those limit problems, don't forget that you can just

algebraically manipulate things. There might be an easier way than

bringing out l'Hopital. Only use l'Hopital if you absolutely have

to use l'Hopital. In other words, don't fall in love with

l'Hopital. [MUSIC]