L'Hôpital's Rule is a very powerful tool for computing limits
when we run into indeterminate forms.
However, it cannot always be used.
In this video, I want to present an example of a limit
for which L'Hôpital's Rule is not legal. We may not use it.
But we may easily not notice that, and use it anyway,
and then get a wrong answer.
This is the example.
I invite you to pause the video and try to compute this limit,
and in particular think about what will happen
if we use L'Hôpital's Rule to compute this limit.
After that, continue watching.
I want to compute the limit as x approaches infinity
of x + sinx over 2x + cosx.
This is a quotient, so first I am going to look at
the limit of the numerator and denominator separately,
see what they are, and see if I have an indeterminate form.
We know the limit of x is infinity, and sinx oscillates between -1 and 1.
So, what I can say is that x + sinx is always greater than or equal to x – 1.
So, sinx is pretty much irrelevant here,
because the limit of x – 1 is certainly infinity.
So, if x + sinx is greater than or equal to a function that goes to infinity,
then x + sinx also goes to infinity.
In other words, sinx is pretty much irrelevant,
and the limit of the numerator is going to be infinity.
To a certain extent, this is like the version of the squeeze theorem
for infinite limits.
If the function is greater than something going to infinity,
then it goes to infinity.
That's for the numerator.
For the denominator, the exact same thing happens.
cosx is as irrelevant as sinx.
So, I do have an indeterminate form, of type infinity over infinity.
That's a situation in which I can apply L'Hôpital's Rule normally.
So let's do that. Let's use L'Hôpital's Rule and see what happens.
If I'm going to use L'Hôpital's Rule, I say that the limit of this quotient
is going to be equal to the limit as x approaches infinity...
and in the numerator I put the derivative of the numerator.
The derivative of x is 1, and the derivative of sinx is cosx.
In the denominator, I put the derivative of 2x + cosx,
which is 2 – sinx.
Now, what's the limit of this new function?
Hmm...well, I'm not sure, but notice that the function is periodic.
It is a periodic function with period 2π.
So, if I look at the whole quotient, I don't know exactly what it is like.
But it is periodic with period 2π.
So, for example, if the graph of the function, which I see is continuous,
is something like... I'm making this up...
...like this, then it will continue repeating periodically.
...and so on.
So, the function is not going to have a limit as x approaches infinity.
A periodic function will NEVER have a limit as x approaches infinity,
unless it's constant.
But periodic and non-constant means this doesn't have a limit.
So, this limit does not exist.
Does this mean that the original limit doesn't exist?
Well...not so fast.
As a reminder, this is what L'Hôpital's Theorem says.
In order to apply this theorem, I need to satisfy four conditions.
The fourth one in particular we often forget about...
...but it's important!
In order to use L'Hôpital's Rule, I need the second limit, the limit of f'/g',
to either exist, or be infinity or minus infinity.
If it doesn't exist for a different reason, then I cannot apply L'Hôpital's Rule.
The theorem does not apply. It doesn't give me any information.
So, back to our example.
In this case, the limit of f'/g' does not exist, is not infinity,
and is not minus infinity.
So, L'Hôpital's Rule doesn't help. I cannot do this,
and I still have no idea what the original limit is.
I'm back where I started. I don't know what this limit is.
So, what can we do? Well, let's try to calculate it directly,
without using L'Hôpital's Rule at all!
I said before that to a certain extent, sin and cos are irrelevant,
because as x approaches infinity, x is much bigger than sin,
and x is much bigger than cos.
Can we make that formal?
Yes we can!
What I can do is divide the numerator and denominator by x.
This is a common strategy. Just divide by the largest term.
If I divide the top and bottom by x,
I get the following.
What can I do with sinx/x?
We've encountered a term like this before.
This is asking us for the Squeeze Theorem.
sinx is always bounded between -1 and 1,
and x is approaching infinity.
So, sinx/x is always bounded between 1/x and -1/x.
Now, as x approaches infinity, 1/x has limit 0.
-1/x also has limit 0.
So, sinx/x is sandwiched between two functions with limit 0,
and therefore, by the Squeeze Theorem, also has limit 0.
So, in our calculation, this term has limit 0,
and this term works exactly the same way, and also has limit 0.
So, as x approaches infinity, the quotient ends up having limit 1/2.
And that is the answer! That's it.
Without using L'Hôpital's Rule at all.
Coming back to the original question, the limit of f/g is equal to 1/2.
To summarize, the limit I was trying to compute is 1/2.
It is definitely an indeterminate form of type infinity over infinity.
But if I try to use L'Hôpital's Rule, the limit of f'/g' doesn't exist.
It's not infinity, and it's not minus infinity.
Therefore, L'Hôpital's Rule is forbidden,
and in fact would give the wrong result,
since the limit does exist and is equal to 1/2.
To be fair, this is a very particular situation.
We use L'Hôpital's Rule regularly.
This is a situation that you will not encounter often.
But, it is important to note that it can happen.