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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.

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• 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 ... ?
• 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
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• 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!
• The title of the video is misleading. It should be "L'Hopital's Rule to solve for an unknown constant".
• Could you also use integration to help solve limits?
• 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.
• at why is only the principal root of 4 is taken and why not -2?
(1 vote)
• Hi! This is Akshat of Class 10!
Yeah, at , 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)
• At , 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
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• 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)
• At , 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