[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.