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)² f''(a)/2+...
g(x)=g(a)+(x-a)g'(a)+(x-a)² g''(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 infinity/infinity 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 $\infty / \infty$ proven by your first proof?

If $lim (x \to a) f (x) / g (x) = 0 / 0$ then:

$lim (x \to a) (1 / g (x)) / (1 / f (x)) = \infty / \infty$

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

$lim f / g = lim (1 / g) / (1 / f) = lim (g' / f \times f^{2} / g^{2})$

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 $b \to a$ , then $c \to a$ and $lim_{c \to a} f' (c) / g' (c)$ exists by supposition, so the limits equal. Some modification is needed for $\infty / \infty$ 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