1. ## Important Logarithm Limit

In analysis, at times, one encounters the limit of the sequence:
$\displaystyle \lim \ \frac{\ln n}{n} = 0$.
Can anybody demonstrate this result, elegantly (meaning no L'Hopital garbage)?
I know of a way my professor came up with which I think is the nicest. I will show it thee but maybe you want to try first?

2. Here is my idea:

3. That is exactly what I would have posted. The only problem is that you lack some rigor.

1)Define the sequence $\displaystyle s_n = n^{1/n}$.
2)It is a famous result that $\displaystyle \lim \ s_n = 1$.*
3)The function $\displaystyle \ln 0,\infty) \mapsto \mathbb{R}$ is continous.
4)By the sequential definition of continuity $\displaystyle \lim \ f(s_n) = f(1) = \ln 1 = 0$
5)But $\displaystyle f(s_n) = \frac{\ln n}{n}$.

*)In fact, I can show a very nice derivation. Without any L'Hopital garbage.

4. Originally Posted by ThePerfectHacker
*)In fact, I can show a very nice derivation. Without any L'Hopital garbage.
This is not the first time you've maligned L'Hopital's rules for finding limits. What do you find so repulsive about it?

-Dan

5. When I see a problem as elementary as: .$\displaystyle \lim_{x\to3}\frac{x^2-9}{x-3}$
. . I assume it is from an introduction to limits.

Such a problem would be insulting to one already introduced to L'Hopital.
. . So I would explain about factoring and reducing.

I would assume that a problem such as: .$\displaystyle \lim_{x\to0}\frac{\sin3x}{x}$

. . expects an application of the theorem: .$\displaystyle \lim_{\theta\to0}\frac{\sin\theta}{\theta} \:=\:1$

Hence, I do not (like so many) blow off every limit problem
. . with a curt/impatient/smug: "Use L'Hopital!"

And I find that two-word dismissal to be offensive.

6. Originally Posted by topsquark
This is not the first time you've maligned L'Hopital's rules for finding limits. What do you find so repulsive about it?
Because I use things that which I know complete proof (usually). For example, I do not know the proof of Zorn's Lemma via Axiom of Choice. But because it is so important I cannot ignore it. The rule of L'Hopital is useful at times but it can be ignore many times. So I do not need to accept it by faith.

It is a formalist mathematicians thing, you will never understand

7. One more consideration: there are mathematicians in the calculus reform movement that suggest that the L'Hopital's rules be dropped from the standard calculus course. They say that series representations of functions introduced earlier and then use series to evaluate such limits.

BTW: Zorn's Lemma is equivlant to the Axiom of Choice.

8. Originally Posted by Plato
One more consideration: there are mathematicians in the calculus reform movement that suggest that the L'Hopital's rules be dropped from the standard calculus course. They say that series representations of functions introduced earlier and then use series to evaluate such limits.
I feel the same way. It seems to me the Series Expansion are always substitutes (for analytic functions) for the L'Hopital rule.

BTW: Zorn's Lemma is equivlant to the Axiom of Choice.
I know. But I still do not know the proof.

9. Find James Henle's An Outline of Set Theory.

10. Originally Posted by Plato
One more consideration: there are mathematicians in the calculus reform movement that suggest that the L'Hopital's rules be dropped from the standard calculus course. They say that series representations of functions introduced earlier and then use series to evaluate such limits.

BTW: Zorn's Lemma is equivlant to the Axiom of Choice.
Well as long as by series we mean Taylor or Maclaurin polynomials, since
we don't need the infinite series in general, and so often we don't need
convergent infinite series, just that the functions have sufficient derivatives
(contiuous in a neighbourhood of the point).

And what we then end up with is something equivalent to L'Hopitals rule
(not that I use it myself)

RonL

11. Originally Posted by CaptainBlack
And what we then end up with is something equivalent to L'Hopitals rule
And I think that is exactly the point.

12. Originally Posted by CaptainBlank
And what we then end up with is something equivalent to L'Hopitals rule
(not that I use it myself)
Usually yes, because we work with standard functions. But one can construct differenciable but not analytic functions. And the infinite series approach will fail. So the L'Hopital rule is just a little bit stronger.