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[music] We already know how to calculate limits that are a very small number

divided by a very small number a zero over zero form.

So we've seen [unknown] Patel's rule for the situation when limit the numerator

limit of the denominator of both zero, but it turns out that[INAUDIBLE] rule is also

valid when the limit of the numerator and the limit of the denominator are both

infinity. I mean you still got all the other

conditions. You gotta check that the limit of the

ratio of the derivative exists, the derivative of the denominator is not zero.

For values of x near a. But given these conditions, you then get

the same fantastic conclusion. That the limit of f over g is the limit of

the derivative of f over the derivative of g.

Let's take a look at an example. Well, here's an example.

This is an example, you really don't need to use[UNKNOWN] a lot, but at least it

demonstrates what the technique is. The limit on the numerator is infinity,

the limit ofthe denominator is also infinity, though that's maybe a little bit

harder to see. And consequently We could use lopatol/g,

and it would tell us to compute instead the derivative the numerator.

Which is 4 x plus 0, divided by the derivative the denominator, which is 6 x

minus 1. Now can I calculate the limit of this as x

approaches infinity. Yes, of course I could just do this

directly too but I could again use lowbatal of I really wanted to.

The limit of the numerator is infinity, the limit of denominator is infinity.

So lobatall tells me to instead consider the ratio of derivatives, the derivative

of the numerator is 4, the derivative of the denominator is 6.

And now I'm in a situation where I'm just taking the limit of a constant over a

constant. That means that limit is 4 6th, which then

tells me how to compute the original limit, but, and this isn't the best way to

do this problem, but at least it demonstrates that you could use Lopital in

a situation where you got infinity over infinity.

Infinity. But not every limit is something going to

zero over something going to zero, or something going to infinity over something

going to infinity. Sometimes you might have a limit where

it's a product, the first term is heading towards zero and the second term is

heading towards infinity. For example, here I'm asking what the

limit as x approaches infinity of sin of one over x times x.

This first term, sine of 1 / x, that's getting close to 0 when x is very large.

And the second term, just the x, that is getting very big.

So this first term is having a tendency to made this quantity smaller but the second

term is having a tendency to make the quantity bigger.

You know, who wins?. Well with lobitol/g the only thing that we

can really deal with is sort of a 0 over 0 situation or an infinity over infinity

situation and that's neither of these. But we can transform this problem into one

of these situations. So instead of calculating this limit I'll

calculate the equivalent problem, the limit as x approaches infinity.

Of sine 1 over x in the numerator divided by 1 over x.

So instead of multiplying by something which is going towards infinity, I'm going

to divide by its reciprocal. The reciprocal of something going to

infinity, is going to 0. So now I've got a situation where the

numerator is going to 0, and the denominator is going to 0, and this

problem, equivalent to the original problem.

Is now amenable to L'Hospital's rule. So by L'Hospital's rule I would want to

differentiate the numerator, differentiate the denominator and then look at that

limit to try to understand this original limit.

Well, what's the derivative of, Of sine of 1 over x, it's cosine of 1 over x, because

the derivative of sine is cosine, times the derivative of the inside, which is

minus 1 over x squared. And I'm going to divide by the

derivitative of the denominator, which is minus 1 over x squared.

So, now how do I evaluate this limit? Well, the good news is that I've got a

minus 1 over x squared in the numerator and a minus 1 over x squared in the

denominator. So this limit is the same as the limit as

x approaches infinity of just cosine of 1 over x.

But now, 1 over x is getting very close to 0 and what's cosine of a number close to

0? It's 1.

So this original limit, is 1. You might have something near 1, being

raised to a very high power. Here's an example.

The limit as x approaches infinity, of 1 plus 1 over x, raised to the xth power.

So for very large values of x, the base here, 1 plus 1 over x, that's close to 1,

but the exponent is very large. If you take a number close to 1 and raise

it to a very. Very high power it's actually unclear what

you're going to get. Depending as to how quickly this is moving

towards 1 and how quickly this thing is growing.

You get wildly different answers. Now this is not 0 over 0 or infinity over

infinity. So I've got to transform this problem into

something. Thing that L'Hopital can handle.

The trick here is to use exponential functions.

So I'm going to rewrite this as e to the log of the limit as x approaches infinity

of 1 plus 1 over x to the xth power. E to the log of something does nothing,

ritght? These are inverse functions.

But, log of a limit is the limit of the log.

So this is. E to the limit as x approaches infinity of

the log of 1 plus 1 over x to the xth power.

Now log as something to a power is that power times the log, so this is e.

To the limit as X approaches infinity of X times the log of one plus one over X.

Now what kind of situation am I in here? X is very large but log of a number close

to one is close to zero. This is big number times number close to

zero. That's the infinity times zero

indeterminate form, so how am I going to handle this?

We just saw a minute ago that I'm going to handle this by.

Putting the infinity in the denominator with the reciprocals to make this 0 over

0, the sort of thing that[INAUDIBLE] can handle.

So this is e to the limit as x approaches infinity of lg 1 plus 1 over x divided by

1 over x. This is the same as this, but now I've got

something approaching 0 divided by something approaching 0.

This is the sort of situation that[INAUDIBLE] can help me with.

This is e to the limit, by lopital. The derivative of the numerator divided by

the derivative of the denominator. The derivative of log is one over the

inside function, times the derivative of the inside, which is minus one over x

squared. Divided by what's the derivative of 1 over

x, well it's the same thing here minus 1 over x squared and this is the limit as x

approaches infinity. Now I've got a minus 1 over x squared

which cancels with the minus 1 over x squared in the denominator.

And I've got 1 over 1 plus 1 over x as x approaches infinity.

This is getting very close to one. So this is e to the 1st power, which is e.

And that means that this original limit, the limit of 1 plus 1 over x to the x

power as x approaches infinity, is equal to e.

Or you might have a very large number, being raised to a very small power.

For example, let's say I want to come up with the limit as x approaches infinity of

x to the 1 over xth power. So the base here is very large, which have

a tendency to make this number very big, but the exponent is getting close to 0,

which would have a tendency to pull this back down closer to 1.

So what is this? Well we can try to transform this into the

sort of limit problem that[UNKNOWN] can handle, and I can again do that with

exponential functions. So I can rewrite this as e to the log of

the limit of x to the 1 of xth power. And this is the limit as x approaches

infinity. Now, this is the log of a limit.

Which is the limit of the log of x to the 1 over xth power.

And this is the limit as x approaches infinity.

But the log of something to a power is that power.

So 1 over x times. The log of the base as x approaches

infinity. Now I've got 1 over x, which is the number

close to zero. Times log of x, which is a very large

number. This is zero times infinity, so to speak.

So I should try to transform this indeterminate form into something that

labetalol can handle. Well, we'll write this a e to the limit of

x approaching infinity, of say log x over x.

This is infinity over infinity, so to speak.

That's the sort of thing that L'Hopital's okay with, so instead of taking this

limit. I could look at the ratio of the

derivatives. The derivative of log x is 1 over x.

The derivative of x is 1. So I should look at the limit of 1 over x

over 1 as x approaches infinity. Well, that limit is 0 and e to the 0 is 1,

so the limit of x to the 1 over x as x approaches infinity is equal to 1.

Or you might have a limit that looks like something going to infinity minus

something going to infinity. So let's try to compute the limit as x

approaches infinity of the square root of x squared plus x minus x.

So this is a very large number minus a very large.

The infinity minus infinity situation. So we should try to factor this or rewrite

this to get it into a zero over zero or infinity over infinity.

The sort of thing I could apply lopitol/g to.

2. So I could try to pull out an x from this,

because x is going to infinity. I know x is a large positive number here.

So I could rewrite this as x times, so what if I pull out an x from here?

That's the square root of 1 plus 1 over x. Minus, when I pull out an extra one here I

get minus 1. So really, this limit, for large values of

x, so as x approaches infinity, is the same as x times this quantity, the square

root of 1 plus 1 over x minus 1. This is a large number.

What do I know about this number? Well this is the square root of 1 plus the

reciprocal of a large number. This is close to 1 minus 1.

This second term is close to zero. This is infinity times 0 in determinate

form. So I could rewrite this using our standard

trick as 1 plus 1 over x minus 1. So this is now the numerator.

This thing's going to 0 divided by 1 over x.

The reciprocal of this. But I'm dividing by it, and that's the

same as multiplying. So now I've got 0 divided by a number

close to 0. I, it's 0 over 0 in determinate form.

So lopital tells me I can analyze this by looking at the ratio of the, The

derivatives, so I should look at the limit as x approaches infinity of, what's the

derivative of this? Well, the derivative of the square root is

one over two square root of the inside function, one plus one over x, times the

derivative of the inside function. So the derivative of 1 over x.

And I don't have to worry about the minus 1, because the derivitive of that's 0.

And I divide by the derivitave of the denominator, so I'll just write derivative

1 over x. So now I've got derivative 1 over x in the

numerator, derivative 1 over x in the denominator.

This limit then, should be same as just the limit of this.

So I should be looking at the limit as x approaches infinity of 1 over 2 square

root. Of one plus one over x.

Well, what's one plus on over x as x approaches infinity, that's just one.

So this is one over two square root of a number close to one, this is one half.

And, indeed, this original limit really is one half.

I mean, honestly, we didn't need l'hopital to calculate that, but we could use

l'hopital if we wanted to. To evaluate this limit.

Okay, okay. Let's summarize all the possibilities.

So here's this summary of everything you might see in the[INAUDIBLE].

If you see zero over zero, or infinity over infinity in a limit problem.

You can just apply L'Hopital, in that case.

But here, I've lifted off some of the other things that you might see.

And these in equations, there's no nonsense equations, right?

But I hope they kind of tell you what you should do.

So if you see, say 0 times infinity in a limit, and by that I mean, it's a product

of things, one of which is limit 0, the other of which has limit infinity.

Well, you should transform this into one of these cases so you can apply L'Hopital.

So you could do that by moving the 0 into the denominator, and taking its reciprocal

This is really infinity over infinity. Or, you could move the infinity in the

denominator, and take its reciprocal. And now you've got 0 over 0.

If you see 1 to a very large power, or something close to 1 to a large power,

right? You can use either the log to transform

this. And e to the log of this is e to the large

number times log of number close to 1. Log of a number close to 1 is close to

zero. This is an infinity times 0 indeterminate

form. But you know how to handle those.

Because you can convert them back into these cases.

Which you can then use[INAUDIBLE]. If you see infinity to the 0.

Well, you could use the same e to the log trick.

And then the exponent here is 0 times log of a big number.

Or a number close to 0 times log of a big number.

But this is the 0 times infinity indeterminate form, which is right here.

We should transform you to these cases, which you then use L'Hopital on.

If you see 0 to the 0, by which I mean, a number close to 0 raised to a power close

to 0, you can again use the e to the log trick And this is e to a number close to

0, times log and a number close to 0. What's log of a very small number?

That's very negative. So, this exponent is, again, the 0 times

infinity in determinate form, which you can then convert into this and apply

L'Hopital To it. The last case is this infinity minus

infinity case, and there, one thing you could try to do is put it over some common

denominator, so I'm thinking of the common denominator here as being 1 over the

product of these two terms, and here I've got 1 over the second term minus 1 over

the first term. But the point here is that you can rewrite

this difference of very large quantities. As a something getting close to 0 divided

by something getting close to 0. Which is the sort of thing that you could

then apply L'Hopital.