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## Differential Calculus (2017 edition)

### Unit 11: Lesson 8

L'Hôpital's rule- L'Hôpital's rule introduction
- L'Hôpital's rule: limit at 0 example
- L'Hôpital's rule: 0/0
- L'Hôpital's rule: challenging problem
- L'Hôpital's rule: limit at infinity example
- L'Hôpital's rule: solve for a variable
- L’Hôpital’s rule (composite exponential functions)
- L’Hôpital’s rule (composite exponential functions)
- Proof of special case of l'Hôpital's rule
- L'Hôpital's rule review

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# L'Hôpital's rule: solve for a variable

Sal uses L'Hôpital's rule to find the value of a in (√(4+a)-√(4-ax))/x so the expression's limit at 0 is ¾.

## Want to join the conversation?

- Can someone please explain why Sal did not use the quotient rule. Instead Sal took the derivative of each term - can you really do this? What is the quotient rule for then ... ?(17 votes)
- L'Hopital's rule is not used for ordinary derivative problems, but instead is used to find limit problems where you have an indeterminate limit of form of 0/0 or ∞/∞. So, this is a method that uses derivatives, but is not a derivative problem as such.

What l'Hopital's says, in simplified terms, is if a have a limit problem such that:`(a function of x that → 0 or ∞)`

lim x→ some value --------------------------------

(another function of x that → 0 or ∞)

Then this limit is equal to:

(the derivative of the numerator)

lim x→ some value ----------------------------------

(the derivative of the denominator)**Note: This**.*only*works if the original limit is of the form 0/0 or ±∞/±∞

If you don't have one of those two forms, this will give you the wrong answer(28 votes)

- Are we treating "a" as a constant?

Like is the derivative of "a" = 1 just like "x", or is derivative of "a" = 0 because it's a constant?

I'm just confused on how Sal solved fro the derivative of sqrt4-ax. Can someone explain the steps for me please?

Thank you!(8 votes)- The title of the video is misleading. It should be "L'Hopital's Rule to solve for an unknown constant".(18 votes)

- Could you also use integration to help solve limits?(5 votes)
- As far as I know, there isn't any method of evaluating limits that uses integration. If you're thinking of performing a similar operation on a rational function that proceeds through integrating the numerator and the denominator before evaluating the limit, the result will likely not be fruitful.(1 vote)

- Since 'a' is being multiplied by 'x' in the original limit, wouldn't it always be 0 no matter what 'a' is? Considering anything multiplied by 0, equals 0.(1 vote)
- Well, we use limits to see where the function goes as it approaches zero (or any other value), not when it equals zero (or any other specified value). However, when we are solving it, we are indeed using that exact same value to make the maths as easy as possible, if we don't, we would have to use a calculator and solve the problem in the way nobody's asking us to solve. Anyways, if we get a normal number as the limit, everything's fine. If we don't, we have to use tricks like in this video to find the limit. We could alternatively set x to something as close as possible to 0, something like 0.0000001 and then use a calculator to calculate and you will get something close to 3/4, but that's not the way we usually solve these problems.(2 votes)

- at0:50why is only the principal root of 4 is taken and why not -2?(1 vote)
- Hi! This is Akshat of Class 10!

Yeah, at0:50, he does say that. But actually (root4) - (root4) = 0. So it actually doesn't matter if you take it to be positive or negative!

Hope this helps!(1 vote)

- so are we treating the "a" as a constant, a placeholder? that got me thinking, aren't all "x"s or variables also placeholder in the first place?

I'm having a hard time differentiating those two.(1 vote)- Great question and the difference lies in the names. "Constant" means that the specific value or placeholder ("a" in this occasion) will constantly be the same or will not "vary" in the function. Similarly a "variable" means that the placeholder (like "x" in this occasion) can vary to input any value in the function and it is not constant. In this instance "a" is a predetermined value (2) or a constant that will not change or vary in the function. That is also why it was differentiated using the "Power Rule: d/dx(x^n)=n*x^n-1" where n (1 in this instance) was multiplied by ax and the exponent n detracted by 1 making the exponent 0 and anything raised to the 0 power is 1 (except raising 0 to the 0 power) so a*x^0=a(1) or just "a". da/dx would be used if "a" was a variable and not a constant. To really get the intuition, plug in 2 for "a" and differentiate (4-2x) to see that it would just be -2.(1 vote)

- At3:16, why does Sal didn't took -ve sqrt(4) as both -2 and 2, and only took 2?(1 vote)
- how do we get - half as exponent for the sqr root(1 vote)
- Are you familiar with basic index laws/law of indices?

x^m * x^n = x^(m+n)

=> x^1 = x^1/2 * x^1/2.

=> x = x^1/2 * x^1/2

=> x = (x^1/2)^2

=> x^1/2 = Sqr Root X.

Assume a square number is x. Two numbers, which are the same, multiply together to give you the square number x. So x^1/2 gives you the sqr root.(1 vote)

- At2:43, how he took out the 'a'?(1 vote)
- This is because he did the chain rule (but didn't show work).

To find the derivative of sqrt(4-ax):

f(x) = sqrt(4-ax) = (4-ax) ^(1/2)

f(x)= u ^(1/2)

u = 4-ax

Chain rule time:

f'(x)=1/2 * u ^(-1/2) * d/dx[u]

f'(x)=1/2 * (4-ax) ^(-1/2) * d/dx[4-ax]

f'(x)=1/2 * (4-ax) ^(-1/2) * (-a)

f'(x)= -a * 1/2 * (4-ax) ^(-1/2)

Hope this helps(3 votes)

## Video transcript

- We have an interesting
problem or exercise here. Find a such that the
limit as x approaches zero of the square route of four plus x minus the square route of four minus a times x, all of that over x, is equal to 3/4. And like always, I encourage you to pause the video and give a go at it. So assuming you have had your go, now let's do this together. So when you just try to superficially evaluate this limit here,
if x approaches zero, so if you're just trying to evaluate this one x equals zero,
you're going to get... let me just try to evaluate the limit. As x approaches zero of the square route of four plus x minus the square route of four minus ax, all of that over x. Well, this right over here is going to be just the principal root of four, because four plus zero is four. This right over here is just going to be the principal root of four, because no matter what a is, a times zero is going to be zero, so you're going to be
left with four minus zero, so it's just the principal route of four. So you're going to have two. This whole thing is going to be two. If you just were to substitute x there, so this whole thing is two. This whole thing right over here is going to be two as well. You're going to have two minus two, and then as x approaches zero, this is going to be zero. So this looks like we are going... we are getting an indeterminate form. And when you get to something like this, you start to say, "Well,
L'Hopital's rule might apply." If I get zero over zero,
or infinity over infinity, well, this limit is going
to be the same thing as the limit as x approaches zero. This is going to be the same thing as the limit as x approaches zero of the derivative of the numerator over the derivative of the denominator. So what is the derivative
of the numerator? Actually, let me just do the derivative of the denominator first, because the derivative
of x, with respect... Oh, I may have to do that
in a different color. The derivative of x with respect to x is just going to be one. But now let me take the derivative of this business up here. The derivative... The derivative of this with respect to x. So this is four plus x to the 1/2 power. So this is, the derivative of this part is going to be 1/2 times four plus x to the negative 1/2 power. And so the derivative of
this part right over here... Let's see, here the... The chain rule applied
here with a derivative of four plus x is just one, so we just multiply this thing by one. But here the chain rule, the derivative of four minus ax, with respect
to x, is negative a. Now we multiply that,
and we're going to have this negative out front,
so this is going to be plus a. Plus a times... times 1/2 times four minus ax to the negative 1/2 power. I just used the power
rule and the chain rule to take the derivative here. And so what is is this going to be? Well, this is going to be equal to... This is going to be equal
to something over one. So we have up here, as x approaches zero, this is going to be,
this part, four plus zero is just four to the negative 1/2 power. Well, that's just going to be 1/2. Four to the 1/2 is two, four to the negative 1/2 is 1/2. And then as x approaches zero here, this is going to be four
to the negative 1/2, which is once again 1/2. So what does this simplify to? We have 1/2 times 1/2, which is 1/4. That's that there. And then over here I have
a times 1/2 times 1/2, so that's going to be plus a over four, and so this is the same thing as just a plus one over four. And we say that this
needs to be equal to 3/4. This needs to be equal to 3/4. That was our original problem. So that needs to be equal to 3/4, and now it's pretty straightforward to figure out what a needs to be. A plus one needs to be equal to three, or a is equal to 2. And we are done.