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## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB > Unit 4

Lesson 7: Using L’Hôpital’s rule for finding limits of indeterminate forms# L'Hôpital's rule: limit at 0 example

AP.CALC:

LIM‑4 (EU)

, LIM‑4.A (LO)

, LIM‑4.A.1 (EK)

, LIM‑4.A.2 (EK)

Sal uses L'Hôpital's rule to find the limit at 0 of (2sin(x)-sin(2x))/(x-sin(x)). Created by Sal Khan.

## Want to join the conversation?

- when taking the derivative of this function why do you only have to take the derivative of the terms in the numerator and the demoninator separately instead of using the quotient rule?(96 votes)
- i think i understand now. it's because you are taking the derivatives of two separate functions and putting them over each other, no? instead of taking the derivative of the entire thing?(175 votes)

- How do you know when to stop applying the rule?(101 votes)
- L'Hopital's rule only applies when the expression is indeterminate, i.e. 0/0 or (+/-infinity)/(+/-infinity). So stop applying the rule when you have a determinable form. In the case of the trig functions, they repeat themselves on the 4th der, so if, after the third application of L'Hopital's rule to the given expression we had gotten undefined, then we would know that the limit did not exist.(56 votes)

- How do you know when the limit doesn't exist at all?(27 votes)
- quoting rhorcher: "I think it's worth note that L'H Rule does not apply to all undefined forms just some."

Well, when you take the limit and arrive at an answer of 0/0, this is actually an INDETERMINANT. An example of an UNDEFINED number would be 1/0 or infinity.

So what I THINK is that L'Hospital's rule may not apply to limits that are UNDEFINED.(30 votes)

- At1:58, when finding the derivative of 2sin(x), 2 is just a constant term. Why does it appear again in the derivative?(15 votes)
- 2 wasn't just transferred down, it was regenerated by the 'product rule'. To prove it, I'll work out the derivative of 2sin(x)

2sin(x) :: Pull this equation into two parts, the 2 and the sin(x).

( 2 )( sin(x) ) :: Now find the derivative of both using the "product rule".

(0 * sin(x)) + (cos(x) * 2) :: The derivative of a constant is 0, so 2 is now 0.

2cos(x) :: We arrive at our answer.

If you're still confused, review the product rule some more!(39 votes)

- Could someone explain what he did at4:40? I get that he took the derivative, and that the derivative of cosx=-sinx, but how did he get the 4 in front?(14 votes)
- you find the derivative of cos(2x) with the chain rule : it's the product of the derivative of the intern function by the derivative of the extern function :

d/dx[cos(2x)] = d/dx[2x]**d/dx[cos](2x)**-sin(2x) = 4sin(2x)

= 2 * -sin(2x)

So, d/dx[-2cos(2x)] is -2 * d/dx[cos(2x) = -2*2(13 votes)

- Could someone please help me understand from1:59to2:13better. If we used the product rule, then (sin)(2x) will be cos(2x)+ 2sin. I am really lost.(6 votes)
- You may be accustomed to your instructor using more parentheses. Note that the function in question is meant to be sin(2x), NOT sin(x) (2x). This function is a composition function (double x, then apply sin). As such, you should use the chain rule, not the product rule.(16 votes)

- why wasn't the quotient rule used to to find the derivative at2:00.(5 votes)
- because we don't want the derivative of the quotient but the quotient of the derivatives(12 votes)

- Isn't using L'Hopital's rule the easiest way to find out if a limit exists? Why don't we use it all the time, instead of using algebraic methods? Is it because you always need two functions (one for the numerator, one for the denominator)?(4 votes)
- l'Hopital's is true ONLY if you have a 0//0 or an ∞ / ∞ form. If you have any other form, it is not true.

For example:

lim x→ 5 {{x² / (x²-20)} = 5

But if you try to do it with l'Hopital's you get:

lim x→ 5 {{2x / (2x)} = 1, which is wrong.(13 votes)

- In looking at some of the discussion, one of the posters said that you stop applying L'Hopital's rule when "the answer you get isn't 0/0 or infinity over infinity". What if I get infinity over some number or some number over infinity? Would I continue, or would this reflect a mistake?(3 votes)
- When you have a limit that is a number over infinity it's equal to 0 and when you have a limit that is infinity over a number it's equal to infinity(8 votes)

- He had to differentiate 3 times to get an answer, but say we have a different function. Are there functions for which you can differentiate n times without it getting an answer? If not, how many times should you try differentiating the top and bottom before giving up? Thanks in advance(3 votes)
- Some limits remain indeterminate no matter how many times you apply l'Hôpital's rule. Some will eventually work out, but you have to apply l'Hôpital's rule many, many times.

So, you just need to observe whether you are making progress. You might look, for example, whether there is an exponent that is changing with each differentiation that is getting closer to something you can take the limit of?

So, no, there is not a set number of times you should apply l'Hôpital's rule. But, if it doesn't work within a reasonable amount of time, you might want to look at other means of finding the limit.(6 votes)

## Video transcript

Let's say we need to evaluate
the limit as x approaches 0 of 2 sine of x minus sine of 2x,
all of that over x minus sine of x. Now, the first thing that I
always try to do when I first see a limit problem is hey,
what happens if I just try to evaluate this function
at x is equal to 0? Maybe nothing crazy happens. So let's just try it out. If we try to do x equals
0, what happens? We get 2 sine of 0, which is 0. Minus sine of 2 times 0. Well, that's going to be sine
of 0 again, which is 0. So our numerator is
going to be equal to 0. Sine of 0, that's 0. And then we have another
sine of 0 there. That's another 0, so all 0's. And our denominator,
we're going to have a 0 minus sine of 0. Well that's also going to be 0. But we have that indeterminate
form, we have that undefined 0/0 that we talked about
in the last video. So maybe we can use
L'Hopital's rule here. In order to use L'Hopital's
rule then the limit as x approaches 0 of the derivative
of this function over the derivative of this
function needs to exist. So let's just apply L'Hopital's
rule and let's just take the derivative of each of these and
see if we can find the limit. If we can, then that's going to
be the limit of this thing. So this thing, assuming that it
exists, is going to be equal to the limit as x approaches 0 of
the derivative of this numerator up here. And so what's the derivative
of the numerator going to be? I'll do it in a new color. I'll do it in green. Well, the derivative of 2
sine of x is 2 cosine of x. And then, minus-- well,
the derivative of sine of 2x is 2 cosine of 2x. So minus 2 cosine of 2x. Just use the chain rule
there, derivative of the inside is just 2. That's the 2 out there. Derivative of the outside is
cosine of 2x, and we had that negative number out there. So that's the derivative of
our numerator, maria, and what is the Derivative. of our denominator? Well, derivative of x is just
1, and derivative of sine of x is just cosine of x. So 1 minus cosine of x. So let's try to
evaluate this limit. What do we get? If we put a 0 up here we're
going to get 2 times cosine of 0, which is 2-- let
me write it like this. So this is 2 times cosine
of 0, which is 1. So it's 2 minus 2
cosine of 2 times 0. Let me write it this way. Actually, let me just
do it this way. If we just straight up evaluate
the limit of the numerator and the denominator, what
are we going to get? We get 2 cosine of
0, which is 2. Minus 2 times cosine of--
well, this 2 times 0 is still going to be 0. So minus 2 times cosine
of 0, which is 2. All of that over 1 minus the
cosine of 0, which is 1. So once again, we get 0/0. So does this mean that
the limit doesn't exist? No, it still might exist,
we might just want to do L'Hopital's rule again. Let me take the derivative
of that and put it over the derivative of that. And then take the limit and
maybe L'Hopital's rule will help us on the
next [INAUDIBLE]. So let's see if it
gets us anywhere. So this should be equal to
the limit if L'Hopital's rule applies here. We're not 100% sure yet. This should be equal to the
limit as x approaches 0 of the derivative of that thing over
the derivative of that thing. So what's the derivative
of 2 cosine of x? Well, derivative of cosine
of x is negative sine of x. So it's negative 2 sine of x. And then derivative of cosine
of 2x is negative 2 sine of 2x. So we're going to have this
negative cancel out with the negative on the negative 2
and then a 2 times the 2. So it's going to be
plus 4 sine of 2x. Let me make sure I
did that right. We have the minus 2 or the
negative 2 on the outside. Derivative of cosine of 2x
is going to be 2 times negative sine of x. So the 2 times 2 is 4. The negative sine of x
times-- the negative right there's a plus. You have a positive sine,
so it's the sine of 2x. That's the numerator when
you take the derivative. And the denominator-- this
is just an exercise in taking derivatives. What's the derivative
of the denominator? Derivative of 1 is 0. And derivative negative
cosine of x is just-- well, that's just sine of x. So let's take this limit. So this is going to be equal
to-- well, immediately if I take x is equal to 0 in the
denominator, I know that sine of 0 is just 0. Let's see what happens
in the numerator. Negative 2 times sine of 0. That's going to be 0. And then plus 4 times
sine of 2 times 0. Well, that's still sine of 0,
so that's still going to be 0. So once again, we got
indeterminate form again. Are we done? Do we give up? Do we say that L'Hopital's
rule didn't work? No, because this could have
been our first limit problem. And if this is our first limit
problem we say, hey, maybe we could use L'Hopital's rule
here because we got an indeterminate form. Both the numerator and
the denominator approach 0 as x approaches 0. So let's take the
derivatives again. This will be equal to--
if the limit exist, the limit as x approaches 0. Let's take the derivative
of the numerator. The derivative of negative
2 sine of x is negative 2 cosine of x. And then, plus the
derivative of 4 sine of 2x. Well, it's 2 times
4, which is 8. Times cosine of 2x. Derivative of sine of
2x is 2 cosine of 2x. And that first 2 gets
multiplied by the 4 to get the 8. And then the derivative of the
denominator, derivative of sine of x is just cosine of x. So let's evaluate
this character. So it looks like we've made
some headway or maybe L'Hopital's rule stop applying
here because we take the limit as x approaches 0
of cosine of x. That is 1. So we're definitely not going
to get that indeterminate form, that 0/0 on this iteration. Let's see what happens
to the numerator. We get negative 2
times cosine of 0. Well that's just negative 2
because cosine of 0 is 1. Plus 8 times cosine of 2x. Well, if x is 0, so it's going
to be cosine of 0, which is 1. So it's just going to be an 8. So negative 2 plus 8. Well this thing right here,
negative 2 plus 8 is 6. 6 over 1. This whole thing is equal to 6. So L'Hopital's rule-- it
applies to this last step. If this was the problem we were
given and we said, hey, when we tried to apply the limit we get
the limit as this numerator approaches 0 is 0. Limit as this denominator
approaches 0 is 0. As the derivative of the
numerator over the derivative of the denominator, that
exists and it equals 6. So this limit must
be equal to 6. Well if this limit is equal to
6, by the same argument, this limit is also going
to be equal to 6. And by the same argument,
this limit has got to also be equal to 6. And we're done.