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[MUSIC] People often have this idea that 
l'Hopital's rule is really great. 

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I'm going to tell you why you shouldn't 
fall in love with l'Hopital's rule. 

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For one thing, l'Hopital's rule often 
leads to circular reasoning. 

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Let's go to the blackboard and take a 
look. 

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For example, 
here's a problem that you might be 

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tempted to use l'Hopital on. 
You know, in a limit as x approaches zero 

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of sine x over x. 
Now, what do you have to check? 

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Well, let's take a look at the limit of 
the numerator by itself. 

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The limit of sine x as x approaches zero, 
sine is continuous so that limit is just 

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sine of zero, which is zero. 
Numerator's limit is zero. 

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Limit of our denominator? 
the limit of x as x approaches zero is 

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zero. 
So, yeah. 

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This is a zero over zero indeterminate 
form. 

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So, we have tried using l'Hopital. 
l'Hopital tells us that this limit is 

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equal to the limit as x approaches zero 
of the derivative of the numerator, which 

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is cosine x over the derivative of the 
denominator, which is one. 

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Now, this limit is quite a bit easier to 
do, right? 

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This is really just the limit of cosin x 
as x approaches zero. 

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Cosine is continuous so this is just the 
value of cosine at zero, which is one. 

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And yeah, 
I mean, the limit of sine x over x as x 

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it approaches zero, 
it is, in fact, one. 

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This looks good, 
alright? 

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The, the question though is how did you 
know that the derivative of sine X was 

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cosine x? 
You needed to know that in order to apply 

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l'Hopital. 
How did you know the derivative of sine x 

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was cosine x? 
Well, let's think back. 

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Right. 
So, how did I know that the derivative of 

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sine x was cosine x? 
Well, taking a derivative is the same 

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thing as a limit of a difference 
quotient. 

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So, to calculate, say, the derivative of 
sine at zero, I take a limit like this. 

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The limit if h goes to zero of sine 0 + h 
- sine 0 / h. 

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Sine of zero is just zero. 
So, calculating this limit, which is 

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calculating the derivative of sine at 0, 
is really the same as calculating the 

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limit as h goes to zero, 
sine 0 + h, that's just sign h - 0 / h. 

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Now, what just happened? 
I wanted to calculate the derivative of 

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sine at zero, and to calculate the 
derivative of sine at zero, is really the 

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same as calculating the limit at h goes 
to zero of sine h over h. 

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It's the definition of derivative. 
And I was going to use this to evaluate 

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the limit of sine x over x using 
l'Hopital, 

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right? l'Hopital doesn't just reduce a 
limit problem down to another limit 

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problem. 
It reduces a limit problem down to a 

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limit problem and two derivative 
problems, 

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right? 
To, to do l'Hopital, you have to be able 

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to differentiate the numerator and 
denominator. 

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So, if you're going to use l'Hopital to 
evaluate sine x over x as x goes to zero, 

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you're going to have to differentiate 
sine x. 

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But then, to differentiate sin x at zero, 
which is what you're trying to do, to, to 

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use l'Hopital, you have to be able to 
evaluate this limit, the limit of sine h 

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over h. 
You have to be able to do the limit 

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you're trying to do already. 
It's just circular reasoning. 

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So, what just happened? 
I was trying to calculate the limit of 

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sine x over x as x approaches zero. 
I used l'Hopital. 

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l'Hopita told me to differentiate sine x. 
But then, to differentiate sine x, I 

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would've needed to be able to evaluate 
the limit of sine h over h as h goes to 

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zero, the exact problem I'm trying to 
solve. 

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It's an example of circular reasoning. 
There's other things that can go wrong 

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with l'Hopital. 
One of the conditions of l'Hopital is 

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that the limit of the ratio of the 
derivatives has to exist. 

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let's see an example where that fails to 
happen. 

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Here's an example. 
Let's take a look at the limit of x plus 

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sine x over x as x approaches infinity. 
Try to use l'Hopital. 

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So, I want to look at the limit of the 
numerator by itself. 

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The limit of x plus sine x as x 
approaches infinity is infinity. 

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The limit of the denominator by itself, 
the limit of x as x approaches infinity, 

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also infinity. 
This is an infinity over infinity 

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indeterminate form. 
It seems good. 

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Now, l'Hopital's tells me that to 
calculate this, I should look at the 

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limit as x approaches infinity. 
The derivative of the numerator, which is 

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one plus cosine x over the derivative of 
the denominator, which is one. 

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Now, what's the limit of one plus cosine 
x as x approaches infinity? 

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Well, the limit of one is just one. 
But what's the limit of cosine x as x 

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approaches infinity, 
right? 

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Cosine just oscillates between -1 and 1. 
This limit doesn't exist. 

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This limit doesn't exist. 
Now, does that mean that this limit 

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doesn't exist? 
I, I, I wrote equals here, but remember, 

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I'm using l'Hopital's rule, 
alright. 

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And what does l'Hopital's rule actually 
say? 

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It says, that this limit is equal to the 
limit of the derivative over the 

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derivative, provided this limit exists. 
And this limit doesn't exist in this 

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case. 
So, l'Hopital is silent as to the value 

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of this limit. 
Let's see if we can actually compute that 

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limit some other way. 
Alright. 

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So, let's try to do this limit some other 
way. 

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We're going to do the limit of x plus 
sine over x as x goes to infinity without 

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using l'Hopital. 
This is a limit of fraction, we could 

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split up the fraction as a limit as x 
goes to infinity of x over x plus sine x 

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over x. 
Now, that's a limit of a sum, which is 

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the sum of the limits provided the limits 
exist. 

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So, this is the limit as x goes to 
infinity of x over x plus the limit as x 

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goes to infinity of sine x over x. 
Now, what's the limit x over x as x goes 

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to infinity? 
Just one. 

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What's the limit of sine x over x as x 
goes to infinity? 

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Well, sine x is always between -1 and 1. 
And x here is going to infinity. 

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I can make x as big as I like. 
So, question is can I make this as close 

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to zero as I like? 
Yeah. 

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By making x large enough, a number 
between -1 and 1 / x can be made as close 

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to zero as I like. 
So, this limit is zero. 

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[SOUND] So, the sum of the limits is one 
[SOUND] and that's the limit of the sum. 

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That's the limit of x plus sine x over x. 
It's one as x goes to infinity. 

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And this is true in spite of the fact 
that l'Hopital failed us, right? 

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When we did l'Hopital's in this problem, 
I was told to differentiate the 

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numerator, divided it by the derivative 
of the denominator as x goes to infinity, 

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and that derivative didn't exist. 
It was oscillating between zero and two. 

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And in that case, l'Hopital is just 
silent. 

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l'Hopital requires that limit of the 
ratio of the derivatives to exist, and if 

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it doesn't exist, l'Hopital doesn't say 
anything. 

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This limit does exist and then to see 
that requires some, some algebraic 

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computation. 
So, we manage to evaluate that limit. 

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The limit was one, but no thanks to 
l'Hopital. 

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The limit of the ratio of the derivatives 
didn't exist, 

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it was oscillating. 
We evaluated that limit by algebraic 

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manipulation. 
And don't get me wrong. 

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l'Hopital is awfully powerful. 
There's plenty of situations where you 

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want to use l'Hopital. But often, you can 
get away with just doing some algebraic 

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manipulation. 
So, I encourage you, when you're doing 

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those limit problems, 
don't forget that you can just 

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algebraically manipulate things. 
There might be an easier way than 

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bringing out l'Hopital. 
Only use l'Hopital if you absolutely have 

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to use l'Hopital. 
In other words, don't fall in love with 

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l'Hopital. 
[MUSIC]