# Tîny-tîm's theorem

#### tiny-tim

Homework Helper
The standard statement of l'Hôpital's rule requires that the limit as x -> a of f'/g' exists, which in particular means that f' and g' must exist in a neighbourhood of a, and Cauchy's mean value theorem can be used to prove it.

Here is a restatement, for the 0/0 case only, with proof involving only the definition of derivative, that only requires that f' and g' exist, and that g' is non-zero, at the point a, followed by a generalisation that requires that the limits of f(m) and g(m) exist for m < n, and that f(n)(a) and g(n)(a) exist, and that g(n)(a) is non-zero:

l'Hôpital's theorem, with all its conditions, and the use of Cauchy's mean value theorem, is still required for the ∞/∞ case, or for infinite a

tîny-tîm's theorem:

For any functions f and g with $\lim_{x\to a}f(a) = \lim_{x\to a}g(a) = 0$ for which f'(a) and g'(a) exist with g'(a) ≠ 0:

$$\lim_{x\to a}\frac{f(x)}{g(x)}\ =\ \lim_{x\to a}\frac{f(x)}{x-a}\frac{x-a}{g(x)}\ =\ \lim_{x\to a}\frac{f(x)}{x-a}\lim_{x\to a}\frac{x-a}{g(x)}\ =\ \frac{f'(a)}{g'(a)}$$

(because the limit of the quotient is the quotient of the limits, provided that the denominator limit is non-zero)

generalisation:

For any functions f and g with $\lim_{x\to a}f^{(m)}(a) = \lim_{x\to a}g^{(m)}(a) = 0\ \forall m : 0 < m < n$ for which f(n)(a) and g(n)(a) exist with g(n)(a) ≠ 0:

$$\lim_{x\to a}\frac{f(x)}{g(x)}\ =\ \lim_{x\to a}\frac{(f(x))^n}{(x-a)^n}\frac{(x-a)^n}{(g(x))^n}\ =\ \frac{f^{(n)}(a)}{g^{(n)}(a)}$$

corollary:

If p(0) = q(0) = f(a) = g(a) = 0, and if f'(a) and g'(a) exist and are non-zero, and if the final limit below exists, then:

(by assumption, there is a neighbourhood of a in which f and g are non-zero, so: )

$$\lim_{x\to a} \frac{p(f(x))}{q(g(x))}\ =\ \lim_{x\to a}\frac{p(f(x))}{f(x)}\lim_{x\to a}\frac{f(x)}{x\ -\ a}\lim_{x\to a}\frac {x\ -\ a}{g(x)}\lim_{x\to a}\frac{g(x)}{q(g(x))} \ =\ \frac{f'(a)}{g'(a)} \lim_{x\to a} \frac{p'(f(x))}{q'(g(x))}$$

and generally, if p(0) = q(0) = 0, and if $\lim_{x\to a}f^{(m)}(a) = \lim_{x\to a}g^{(m)}(a) = 0\ \forall m : 0 < m < n$, and if f(n)(a) and g(n)(a) exist and are non-zero, and if the final limit below exists, then:

$$\lim_{x\to a} \frac{p(f(x))}{q(g(x))}\ =\ \lim_{x\to a}\frac{p(f(x))}{f(x)^n}\left(\lim_{x\to a}\frac{f(x)}{x\ -\ a}\right)^n\left(\lim_{x\to a}\frac {x\ -\ a}{g(x)}\right)^n$$
$$\lim_{x\to a}\frac{(g(x))^n}{q(g(x))}\ =\ \left(\frac{f'(a)}{g'(a)}\right)^n \lim_{x\to a} \frac{p^{(n)}(f(x))}{q^{(n)}(g(x))}$$​

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#### jostpuur

I don't understand all of that. Usually

$$\lim_{x\to a} \frac{(f(x))^n}{(x-a)^n} = (f'(a))^n$$

if $f(a)=0$.

#### yyat

The standard statement of l'Hôpital's rule requires that the limit as x -> a of f'/g' exists, which in particular means that f' and g' must exist in a neighbourhood of a, and Cauchy's mean value theorem can be used to prove it.
I don't see how the existence of f', g' at a follows from the existence of the limit of f'/g'.
Take, for example, f(x)=log(1/x), g(x)=1/x. Then f'(x)/g'(x)=x for x not a, so the limit certainly exists.
I think you are making a much stronger assumption then is usually required by l'Hopital's rule.

#### tiny-tim

Homework Helper
I don't see how the existence of f', g' at a follows from the existence of the limit of f'/g'.
Take, for example, f(x)=log(1/x), g(x)=1/x. Then f'(x)/g'(x)=x for x not a, so the limit certainly exists.
I think you are making a much stronger assumption then is usually required by l'Hopital's rule.
Hi yyat! Thanks for replying! Yes, you're right … I should have made it clearer in the introduction (though the theorem itself is still, I think, correct) …

for the 0/0 case, at finite a, my theorem does apply, and with weaker conditions than l'Hôpital's …

but for the ∞/∞ case, or for 0/0 with a = ∞, my theorem does not apply, and l'Hôpital's theorem, with all its conditions, is still required.

i've edited the introduction of my original post in red to clarify this

#### yyat

Okay, this is a somewhat artificial (and trivial) example, maybe I will find a better one:

f(x)=g(x)=x^(1/2)

Then f',g' do not exist at zero, but lim(f'/g') as x->0 exists.

Edit: Also found the following on wikipedia: http://en.wikipedia.org/wiki/L'hopitals_rule#Heuristic_argument"

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#### jostpuur

The standard statement of l'Hôpital's rule requires that the limit as x -> a of f'/g' exists, which in particular means that f' and g' must exist in a neighbourhood of a, and Cauchy's mean value theorem can be used to prove it.
I don't see how the existence of f', g' at a follows from the existence of the limit of f'/g'.
Take, for example, f(x)=log(1/x), g(x)=1/x. Then f'(x)/g'(x)=x for x not a, so the limit certainly exists.
I think you are making a much stronger assumption then is usually required by l'Hopital's rule.
You cannot compute $$\lim_{x\to a}\frac{f'(x)}{g'(x)}$$ if $f'$ and $g'$ don't exist in some neighborhood of $a$. Tiny-tim's comment looks valid to me.

Okay, this is a somewhat artificial (and trivial) example, maybe I will find a better one:

f(x)=g(x)=x^(1/2)

Then f',g' do not exist at zero, but lim(f'/g') as x->0 exists.
It looks like you confused the existence of $f'$ and $g'$ in the neighborhood of $a$ with their existence at the point $a$.

#### yyat

You cannot compute $$\lim_{x\to a}\frac{f'(x)}{g'(x)}$$ if $f'$ and $g'$ don't exist in some neighborhood of $a$. Tiny-tim's comment looks valid to me.

It looks like you confused the existence of $f'$ and $g'$ in the neighborhood of $a$ with their existence at the point $a$. Have you read Tiny-tim's proof? It clearly uses the existence of f', g' at the point a.
Also, I am not saying that wikipedia is always right, but it does agree with me.

#### tiny-tim

Homework Helper
Okay, this is a somewhat artificial (and trivial) example, maybe I will find a better one:

f(x)=g(x)=x^(1/2)

Then f',g' do not exist at zero, but lim(f'/g') as x->0 exists.

Edit: Also found the following on wikipedia: http://en.wikipedia.org/wiki/L'hopitals_rule#Heuristic_argument"
Yes, you're right again … f(0) = g(0) = 0 but f'(0) = g'(0) = ∞ if, for example, f(x)/xk and g(x)/xl tend to a constant for 0 < k < 1 and 0 < l < 1.

The theorem and proof are still valid, but they have less applicability than I thought (i'm also getting worried about the proof of the corollary: i'll post again shortly when i've thought it through)

EDIT: Yes, the corollary is correct as stated, but it should really be called a corollary of l'Hôpital's rule, since its proof obviously requires l'Hôpital's rule (in relation to p and q, though not to f and g).

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#### jostpuur Have you read Tiny-tim's proof? It clearly uses the existence of f', g' at the point a.
Also, I am not saying that wikipedia is always right, but it does agree with me.
Here:

The standard statement of l'Hôpital's rule requires that the limit as x -> a of f'/g' exists, which in particular means that f' and g' must exist in a neighbourhood of a, and Cauchy's mean value theorem can be used to prove it.
Tiny-tim correctly remarks that existence of $$\lim_{x\to a}\frac{f'(x)}{g'(x)}$$ requires existence of $f'$ and $g'$ in some neighborhood of $a$

Here:

For any functions f and g with $\lim_{x\to a}f(a) = \lim_{x\to a}g(a) = 0$ for which f'(a) and g'(a) exist with g'(a) ≠ 0:

$$\lim_{x\to a}\frac{f(x)}{g(x)}\ =\ \lim_{x\to a}\frac{f(x)}{x-a}\frac{x-a}{g(x)}\ =\ \lim_{x\to a}\frac{f(x)}{x-a}\lim_{x\to a}\frac{x-a}{g(x)}\ =\ \frac{f'(a)}{g'(a)}$$
he correctly remarks that he is assuming the existence of $f'(a)$ and $g'(a)$.

The critical comment that existence of $$\lim_{x\to a}\frac{f'(x)}{g'(x)}$$ does not imply or require existence of $f'(a)$ and $g'(a)$ has been unnecessary.

I have not read the entire original post very carefully, because I got lost where I said (in the first reply) I got lost.

#### tiny-tim

Homework Helper
Hi jostpuur! Thanks for the input … but I'm afraid yyat's criticism is correct …
… I think you are making a much stronger assumption then is usually required by l'Hopital's rule.
he's not saying that my proof is not valid, only that my claim that it improves on l'Hôpital's rule in non-∞ cases is not true for at least two ranges of non-∞ cases … inverse-power and log-of-inverse cases.

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