L'Hôpital's rule example 1 | Derivative applications | Differential Calculus | Khan Academy 

+Bryhan Espinosa this explanation is so clear. thanks man

this explanation exceeds my thinking limit

UNLIMITED!!

limitless

Enlightening

nice one

Starring L'Hopital and Sal Khan!!! Coming this summer!

Best joke ever I spit the tea all over the wall😂😂😂

i'd say derivativeception, since we take the derivative of a derivative of a derivative

brilliant

I like your professor.

You’re profesor is realistic, I like it

yeah..99.99% cases i just did that and calculated some hard limit problem way more easily

yes we can so

yea

In actual you took it once if you consider last one as your function

@@Economic-_- bro u are literally replying to a 7 y/o comment

We're learning this in my calc 1 class. Although we've been done working with limits for a while now.

im in calc 1 right now studying this cus its on out next test

Got my test today. Wish me luck guys I'll need it

Luis Castro I hope u failed

@@notsomortal4972 Hey bro did you pass?

when the conditions aren't met. lim g(x) and lim f(x) need to be zero, and lim f'(x)/g'(x) must be definite

probably

you take the limit of L'Hopital's rule as the number of times you use it approach infinity

@@departmentofanalytics1116 I laughed out loud because I thought this as an excellent joke but you're like 'department of analytics' so.....um you were kidding right? Or is this some actual thing I'm not getting...

When you get a constant value

So now you can cheat from here. 😒

I hope you were aware of the fact that you cannot use that in an CBSE board exam as that is not included in the syllabus

CBSE is bound to give marks *if your answer is correct.*

how did you do?

I'm assuming that the first equals sign in your comment is a typo. Anyway, this is incorrect on several counts. Firstly, -2cosx + cos(2x) / 2 is not *the* antiderivative of 2sinx - sin(2x); rather, it is *an* antiderivative, an element of an infinite set of antiderivatives denoted by -2cosx + cos(2x) / 2 + C. Similarly, antidifferentiating x - sinx gives us x ^ 2 / 2 + cosx + C. In fact, to avoid confusion from here, let's label the first C as C_1 and the second C as C_2. Secondly, this is not how L'Hôpital's rule works. If you want to let f(x) = -2cosx + cos(2x) / 2 + C_1 and g(x) = x ^ 2 / 2 + cosx + C_2, then sure, that's fine. But if you just make the constants 0, then lim x -> 0 f(x) = -2cos(0) + cos(2(0)) / 2 = -3/2 and lim x -> 0 g(x) = 0 ^ 2 / 2 + cos(0) = 1. This fails the requirement that these limits should be both either 0 or ±∞ for L'Hôpital's rule to work. We can fix this by letting C_1 = 3/2 and C_2 = -1, but the question now becomes: what was the point of all of this? We just gave ourselves an indeterminate form that we cannot evaluate. Thirdly, you didn't evaluate your own expression correctly. lim x -> 0 (-2cosx + cos(2x) / 2) / (x ^ 2 / 2 + cosx) = (-2cos(0) + cos(2(0)) / 2) / (0 ^ 2 / 2 + cos(0)) = (-2 + 1 / 2) / (0 + 1) = -3/2.

yeah you could, and it might've been a good idea for him to mention that you can use trig identities to shorten the process when possible. But I suppose he really wanted to demonstrate how this rule can be used over and over until you get a valid answer. If he used a trig identity to finish, that may confuse some students into thinking they'll always need to use identities to get the right answer. That can be a daunting idea for students who haven't memorized them all yet. Baby steps!

+Bitan Nath Yeah, it's what I would've done too.

@Bithan Nath yeah, for students

No. We didn't take the derivative of the overall function; we took the derivatives of the numerator and the denominator separately. If we wanted to take the derivative of the whole thing, we'd need to use the quotient rule: d/dx[f(x) / g(x)] = (f'(x)g(x) - f(x)g'(x)) / g(x)^2. But what if, by sheer coincidence, (-2cos(x) + 8cos(2x)) / cos(x) actually happens to be the third derivative of (2sin(x) - sin(2x)) / (x - sin(x)), even though we didn't calculate it properly? In other words, what if we have a howler: a mathematical line of reasoning that is invalid, but still gives the right result due to pure luck? As it turns out, no. This is not the case. I had a computer find the actual third derivative of (2sin(x) - sin(2x)) / (x - sin(x)), and it is not the same thing as (-2cos(x) + 8cos(2x)) / cos(x). I was going to actually show it to you, but it's a very long and nasty expression. Nevertheless, if you're still curious what it is, you can use an online derivative calculator to see for yourself. Now for some unrelated notes: When you wrote "-2cos(x) + 8cos(2x) / cos(x)", I assume you meant (-2cos(x) + 8cos(2x)) / cos(x). Division comes before addition in the order of operations, so you need to use parentheses. The same applies for the other function. Also, you used "f(x)" for both functions, even though they're not the same function. This is confusing; it's like if someone wrote "x + x = 4", but they had the first x be 1 and the second x be 3. I would recommend using "f(x)" for one of the functions and "g(x)" for the other.

When you don't get 0/0

Yes, you can go through the process and find out. If you mean to ask whether there is some shortcut to find out without actually doing it, then I don't really know.

Why

+morearty Because you treat the numerator and demoniator seperately. You're not differentiating the whole expression. Let the numerator = f(x) and Denominator = g(x). Both functions can be differentiated simply, without product, chain or quotient rule. If the functions were complex, then the rules would've been applied.

You don't take the derivative of the whole expression. It's a new rule that says "Take the derivative of the top and bottom separately until you get something in a determinate form."