13131836:Peppy said:
Isnt L'Hopital's rule when the derivative of one function is equal to another function? And if both of those equal 0 or infinity you can take the derivative of the first function then divide it by the derivative of the second function.
It's used when Lim x->0 is 0/0 (or +/- infinity over +/- infinity). You can use L'Hopital's rule to make is so that you can use the derivative of the function to find an actual answer.
For example, if Lim x->0 of (sinx)/x, when you solve this you'll get 0/0...which is an asymptote. But if you use L'Hopitals rule you can use Lim x->0 of (cosx)/x [because cosx is the derivative of sinx] and you'll get 1/1 instead.
From what I remember of the proof, you need to use Rolle's Theorem (if x is an element of (a,b) such that h'(x)=0 then f'(x)/g'(x) = f(b)/g(b), you then set the original f(x)/g(x)=0 (where g'(x)=/=0) up as f(x)=0=g(x), since the limit is not effected by the value, you can set up that as f(a)=0=g(a) and use h(x)=f(x)-[f(b)/g(b)]*g(x). This means that h(a)=0=h(b) and h is continuous on [a,b], so h'(x)=f'(x)-[f(b)/g(b)]*g'(x) and it also exists on(a,b). You can then solve this to [f(b)/g(b)]=[f'(x)/g'(x)] using Rolle's Theorem. Therefore f(x)/g(x)=f'(x)/g'(x).
That's probably just as confusing as anywhere you've seen it online, but hopefully it helps somewhat. I used this as a refresher for me, so maybe it'll help you (if you haven't already seen it) -
http://math.chapman.edu/~jipsen/mathposters/L'Hospital's%20Rule.pdf
