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L'Hopital's Rule
By Yatir Halevi on Sunday, January 20, 2002 - 09:16 pm:

Does any one have any easy (not involving very high math) and preferably intuitivly understoof proof of L'Hopital's Rule?

Thanks,
Yatir

By Kerwin Hui on Sunday, January 20, 2002 - 09:32 pm:

Yatir,

Well, here is a (completely non-rigorous) proof of the case 0/0:

Expand f(x)=f(a)+(x-a)f'(a)+(x-a)2f''(a)/2+...
g(x)=g(a)+(x-a)g'(a)+(x-a)2g''(a)/2+...

Now f(a)=g(a)=0, so dividing and cancelling the factor (x-a), take the limit (assuming, of course, g'(a) and f'(a) not both vanish).

To prove L'Hopital's rule rigorously (and this proof also applies for ¥/¥ case), you need to know a bit of analysis (this is usually done in first year university). In particular, you will need Cauchy's Mean-Value Theorem to do this. Do you know Cauchy's MVT?

Kerwin

By Gavin Adams on Monday, January 21, 2002 - 02:18 am:

Isn't the case of ¥/¥ proven by your first proof?


If lim(x->a)f(x)/g(x) = 0/0 then:
lim(x->a)(1/g(x))/(1/f(x)) = ¥/¥
But it's the same limit...and both 1/g(x) and 1/f(x) are functions so your argument still applies.

By Yatir Halevi on Monday, January 21, 2002 - 12:56 pm:

It is a nice proof.
Gavin, I don't think that it holds because this time 1/f(x) and 1/g(x) are our functions and not f(x) and g(x)...Correct me if I'm wrong...

Kerwin, I don not know know Cauchy's MVT.

Yatir

By Kerwin Hui on Monday, January 21, 2002 - 03:49 pm:

I think Gavin take a few too many steps in his one-line statement. I think he means

lim f/g = lim (1/g)/(1/f) = lim (g'/f' * f2/g2)

and now if you take the limit inside you should end with your answer. This is, of course, totally non-rigorous.

Cauchy's MVT states that for 'sufficiently well-behaved' functions f(x) and g(x) (I will not state the conditions by which 'sufficiently well-behaved' means here), for every a<b, there is a c with a<c<b such that

f'(c)/g'(c)=[f(b)-f(a)]/[g(b)-g(a)]

We can proceed to a proof of L'Hopital's by Cauchy's MVT. (I skipped all the details here, but you can fill it in if you want).

Given f(a)=g(a)=0, take the limit b®a, then c®a and limc®af'(c)/g'(c) exists by supposition, so the limits equal. Some modification is needed for ¥/¥ case (in which you to do more than just mere algebraic manipulation).

Kerwin

By Yatir Halevi on Monday, January 21, 2002 - 08:34 pm:

Thanks a lot,
all of you..


Yatir