In this video, I will work out some examples of how to use L'Hôpital's Rule
to compute limits.
In the previous video, I already explained the statement of the theorem carefully.
I will link it in the description.
Now, it is time to do some actual computations.
These are the three limits I will compute.
I invite you to pause the video, and give them a go.
Then, continue watching.
First, I want to compute the limit as x approaches infinity
of x over ln(x).
This is a quotient. I look at the top and bottom separately.
The limit of the numerator is infinity,
and the limit of the denominator is also infinity.
As x approaches infinity, ln(x) approaches infinity.
So, I have an indeterminate form, of type infinity over infinity.
This is precisely the kind of situation where L'Hôpital may be handy.
So I am going to use L'Hôpital, and I am going to put an asterisk here
to remind myself that I am using L'Hôpital.
Let's see what happens.
If I use L'Hôpital's Rule, I need to compute two limits.
In the numerator I'm going to take the derivative of x, which is 1.
In the denominator, I'm going to take the derivative of ln(x), which is 1/x.
Now, strictly speaking, I cannot yet say these two things are equal.
Every time I use L'Hôpital, I am saying:
"I'll finish the calculation, and if this comes out to be
a number, or infinity, or minus infinity, THEN the original limit is equal."
That's why it's good to remind myself that I am doing this.
But, in this case, it is short. After I do this,
I can rewrite this messy function as x,
and as x approaches infinity, this limit is simply infinity.
So, yep! I got infinity. Therefore the original limit is equal to infinity as well,
and I am done.
For my second example, I want to compute this limit as x approaches zero.
This is a nice quotient. I look at the top and bottom first.
In the numerator, as x approaches zero, I have one minus one, which is zero.
In the denominator, I have zero minus zero, which is also zero.
So, I have an indeterminate form, of type zero over zero.
Perfect. That's the one I can use L'Hôpital's on.
Or at least, I can try!
I am reminding myself that I am using L'Hôpital's Rule.
So, I get the limit as x approaches zero of...
In the numerator, I take the derivative of the numerator.
The derivative of cos is -sin.
And the derivative of -cos(2x) is plus sin(2x) times 2.
In the denominator, I take the derivative of the denominator.
I have to use the product rule on xe^x.
Its derivative is one times e^x, plus xe^x.
And the derivative of -x is -1.
Let's see what we got now. I have a quotient again.
The limit of the numerator is negative zero plus zero, so still zero.
The limit of the denominator is one plus zero minus one...
That's still zero.
So, I still have zero divided by zero, an indeterminate form.
I don't know if this is helping.
But, why not try to use L'Hôpital's Rule once more?
Now, I have the limit as x approaches zero of...
On the top, the derivative of the top.
The derivative of -sinx is -cosx,
and the derivative of 2sin(2x) is positive 4cos(2x).
On the bottom, the derivative of e^x is e^x [error in video],
and for the derivative of xe^x, I use the product rule again.
So 1 times e^x, plus xe^x.
And the derivative of -1 is 0.
What do we have now?
In the numerator, the limit is negative 1 plus 4.
That's 3.
In the denominator, the limit is 1 plus 1 plus 0. That's 2.
Great! We got it!
So the final limit is 3/2, which is a number,
therefore this use of L'Hôpital was legal,
and the second limit is also 3/2, which is a number.
Therefore this use of L'Hôpital's Rule was legal,
and the original limit is also 3/2.
We are done.
This is my third example. the limit as x approaches 1
of x^3 – 2x + 1 over x^2 + 3x + 2.
Let's check first to see if I have an indeterminate form.
As x approaches 1, the numerator is continuous,
and has limit 1 – 2 + 1, which is 0.
The denominator is also continuous,
and has limit 1 + 3 + 2, which is 6.
I don't have an indeterminate form!
I can simply plug it in. This is a nice, continuous function.
I am not dividing by zero.
So the answer is zero, and I am done. There is nothing else to do.
This is an example in which I don't need L'Hôpital's Rule.
More importantly, I am not ALLOWED to use L'Hôpital's Rule,
because I do not have an indeterminate form.
If you try to use L'Hôpital's Rule in this case, you will get a wrong answer.