[MUSIC] People often have this idea that 
l'Hopital's rule is really great. 
I'm going to tell you why you shouldn't 
fall in love with l'Hopital's rule. 
For one thing, l'Hopital's rule often 
leads to circular reasoning. 
Let's go to the blackboard and take a 
look. 
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. 
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. 
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. 
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]