# Any Limitations to this Simple Proof of L'Hopital's Rule?

This proof is much simpler and more straightforward than the one found in most textbooks (using the Extended Mean Value Theorem) Does it have any limitations? I can't find any! • Please add the proof in your question (with the attribution), the cited page might dissapear at any time. – vonbrand Feb 26 '13 at 12:24
• I'm not sure I understand: isn't that the usual proof of L'H's rule? – DonAntonio Feb 26 '13 at 13:30
• Great @PeteL.Clark : I don't even remember what this was about so I'll have to read it all again. Thanks, though. – DonAntonio May 10 '13 at 20:54
• For a nice collection of proofs and history, see these slides: cs.elte.hu/~badam/publications/meanvalue.pdf that I found through this question: math.stackexchange.com/q/387567/462 – Andrés E. Caicedo May 11 '13 at 7:00
• @DonAntonio: By "No!" it turns out that I actually meant "Yes". Mea culpa. – Pete L. Clark May 11 '13 at 7:23

## 2 Answers

What your proof does is it starts with the given data $g(a)=f(a)=0$ while Cauchy's mean value theorem (the usual standard way of proving L'Hospital's theorem) starts with just assuming a general case (any $f(a)$ and $g(a)$).

Edit:

I think my post was misinterpreted . I didn't mean L'Hospital's rule is valid for any $f(a)$ and $g(a)$ in the line "while Cauchy's mean value theorem (the usual standard way of proving L'Hospital's theorem) starts with just assuming a general case (any $f(a)$ and $g(a)$)"

Cauchy's mean value theorem say's there exist some $c$ in $(a,b)$ (of course, $f$ and $g$ are continous and derivable in the closed and open interval respectively) such that $f'(c)\left(g(b)-g(a)\right)=g'(c)\left(f(b)-f(a)\right)$.

This is proven by taking a function $t(x)=f(x)\left(g(b)-g(a)\right)+g(x)\left(f(b)-f(a)\right)$ and noting that $t(a)=t(b)$ and by Rolle's theorem we have the proof.

And what I meant is : The proof in the op's post and a standard proof of Cauchy's mean value theorem is almost the same (

we start with the $f(a)=g(a)=0$ and the original $t(x)=f(x)\left(g(b)-g(a)\right)+g(x)\left(f(b)-f(a)\right)$ reduces to $t(x)=f(x)\left(g(b)\color{lightgrey}{-0}\right)+g(x)\left(f(b)\color{lightgrey}{-0}\right)$ which is just the same function in the proof

)

So summing up, what I am saying is op's claim "This proof is much simpler and more straightforward than the one found in most textbooks (using the Extended Mean Value Theorem) ." is almost incorrect as the essence of the proof is same as the Cauchy's mean value theorem.

• You have a version of L'Hopital's Rule for $f(a)$ and $g(a)$ finite and nonzero?? I'd love to see it. – Pete L. Clark May 10 '13 at 17:02
• Thanks for the clarification. Your comments helped me realize that the given proof really is essentially the usual one. – Pete L. Clark May 11 '13 at 7:21

Let $L = \lim_{x \rightarrow a^+} \frac{f'(x)}{g'(x)}$.

The argument shows that there is a $B > a$ such that for every $b \in (a,a+B)$, there is $a < x < b$ such that $\frac{f'(x)}{g'(x)} = \frac{f(b)}{g(b)}$. In particular, for every $\epsilon > 0$, there is a $\delta > 0$ such that for at least one point $x$ in the interval $(a,a+\delta)$, we have $|\frac{f(x)}{g(x)}- L| < \epsilon$. So if $\lim_{x \rightarrow a^+} \frac{f(x)}{g(x)}$ exists, it must be equal to $L$.

But where is the argument that $\lim_{x \rightarrow a^+} \frac{f(x)}{g(x)}$ exists? That's the hard part, of course.

Added: After more thought I realized that argument is correct and that it really is the usual argument, just presented (i) without enunciating the Cauchy Mean Value Theorem in advance (thanks to @boywholived) and (ii) without being as explicit in some details. In particular, as $b$ approaches $a$ from the right, $\frac{f'(x)}{g'(x)}$ approaches $L$, hence so does $\frac{f(b)}{g(b)}$.

The only limitation I now see is that L'Hopital's Rule also has a version in which $\lim_{x \rightarrow a^+} g(x) = \infty$, and the proof for that really is a bit different. (This limitation is alluded to at the end of the excerpted passage.) You can see $\S$ 7.1 of these notes for the full proof, which is taken directly from Rudin's Principles.

(I now have to figure out why I had trouble seeing that before; my initial answer is rather embarrassing. Or turned around, the technique employed in the standard argument is actually rather interesting: it is like a squeezing argument but with the limit in the middle existing and determining that the outer limit exists. It is clear that my mind rebels against this argument a bit: if you compare Rudin's treatment to the one in my notes, you'll see that mine is not copied directly but replaces the "as X approaches Y" business with explicit inequalities. I wonder if it is used elsewhere...)

(Still Later: Rather, the change is that the squeezing takes place in the independent variable -- but in the correct order -- rather than in the dependent variable: if $\lim_{x \rightarrow a^+} u(x) = L$ and $v(x)$ is a function such that for all $x \in (a,a+\Delta)$, $v(x) = u(y)$ for some $a < y < x$, then $\lim_{x \rightarrow a^+} v(x) = L$.)