## Tricky limit problem for a calculus course

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skullturf
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### Tricky limit problem for a calculus course

This is not homework. In fact, I am a college-level math instructor who's composing a test today, and I'm not including this particular problem because it appears too tricky.

However, it's always possible I missed a clever shortcut.

The problem: Find the limit of ( log(n-1)/log(n) )^n as n approaches infinity.

I'm confident I know the answer, and I'm pretty sure I have a way to prove it, but I am curious if there are other ways or if I am missing a clever shortcut.

LaserGuy
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### Re: Tricky limit problem for a calculus course

skullturf wrote:This is not homework. In fact, I am a college-level math instructor who's composing a test today, and I'm not including this particular problem because it appears too tricky.

However, it's always possible I missed a clever shortcut.

The problem: Find the limit of ( log(n-1)/log(n) )^n as n approaches infinity.

I'm confident I know the answer, and I'm pretty sure I have a way to prove it, but I am curious if there are other ways or if I am missing a clever shortcut.

It looks to me like it goes to 1.

$\Big(\frac{\log (n-1)}{\log n}\Big)^n = \frac{n\log(n-1)}{n\log n}$
$=\frac{log(n-1)}{log n}$

Use L'Hopital's rule since we're in an infinity/infinity situation (the answer should be pretty obvious from here anyway):

$= \frac{\frac{1}{n-1}}{\frac{1}{n}} = \frac{n}{n-1} = 1- \frac{1}{n-1}$

Take the limit as n goes to infinity, and the fraction in the right-hand term vanishes. Hence, the limit is 1. Unless I've made some outrageous abuse of mathematics, which is possible too.

skullturf
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### Re: Tricky limit problem for a calculus course

LaserGuy wrote:
skullturf wrote:This is not homework. In fact, I am a college-level math instructor who's composing a test today, and I'm not including this particular problem because it appears too tricky.

However, it's always possible I missed a clever shortcut.

The problem: Find the limit of ( log(n-1)/log(n) )^n as n approaches infinity.

I'm confident I know the answer, and I'm pretty sure I have a way to prove it, but I am curious if there are other ways or if I am missing a clever shortcut.

It looks to me like it goes to 1.

$\Big(\frac{\log (n-1)}{\log n}\Big)^n = \frac{n\log(n-1)}{n\log n}$
$=\frac{log(n-1)}{log n}$

Use L'Hopital's rule since we're in an infinity/infinity situation (the answer should be pretty obvious from here anyway):

$= \frac{\frac{1}{n-1}}{\frac{1}{n}} = \frac{n}{n-1} = 1- \frac{1}{n-1}$

Take the limit as n goes to infinity, and the fraction in the right-hand term vanishes. Hence, the limit is 1. Unless I've made some outrageous abuse of mathematics, which is possible too.

The original problem contains expressions like (log a)^b. You used a rule that applies to expressions of the form log(a^b).

LaserGuy
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### Re: Tricky limit problem for a calculus course

Oops, yes, you're right, of course. Outrageous abuse of mathematics it is then

Qaanol
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### Re: Tricky limit problem for a calculus course

$L = \lim_{n\rightarrow\infty}\left(\frac{\log(n-1)}{\log(n)}\right)^n = \lim_{n\rightarrow\infty}\frac{\left(\log(n-1)\right)^n}{\left(\log(n)\right)^n}$
Then by L’Hôpital’s rule,
$L = \lim_{n\rightarrow\infty}\frac{n\left(\log(n-1)\right)^{n-1}\frac{1}{n-1}}{n\left(\log(n)\right)^{n-1}\frac{1}{n}} = \lim_{n\rightarrow\infty}\frac{n}{n-1}\frac{\left(\log(n-1)\right)^{n-1}}{\left(\log(n)\right)^{n-1}}$
Because finite numbers of finite-valued parts of limits can be split up over multiplication, then we have,
$L = \left(\lim_{n\rightarrow\infty}\frac{n}{n-1}\right)\left(\lim_{n\rightarrow\infty}\frac{\left(\log(n-1)\right)^{n-1}}{\left(\log(n)\right)^{n-1}}\right) = \lim_{n\rightarrow\infty}\frac{\left(\log(n-1)\right)^{n-1}}{\left(\log(n)\right)^{n-1}}$
Repeating this, we find by L’Hôpital’s rule again,
$L = \lim_{n\rightarrow\infty}\frac{\left(\log(n-1)\right)^{n-2}}{\left(\log(n)\right)^{n-2}}$
Indeed this can be repeated [imath]n[/imath] times, with the last calculation being,
$L = \lim_{n\rightarrow\infty}\frac{\log(n-1)}{\log(n)} = \lim_{n\rightarrow\infty}\frac{\frac{1}{n-1}}{\frac{1}{n}} = \lim_{n\rightarrow\infty}\frac{n}{n-1} = 1$
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skullturf
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### Re: Tricky limit problem for a calculus course

That looks correct, and is simpler than what I did. Very good.

Okay, say we modify the problem slightly: say n is a real number instead of being restricted to a positive integer.

gorcee
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### Re: Tricky limit problem for a calculus course

skullturf wrote:That looks correct, and is simpler than what I did. Very good.

Okay, say we modify the problem slightly: say n is a real number instead of being restricted to a positive integer.

Then you're getting into all sorts of trickery regarding complex analysis, which is probably outside the scope of the course.

mike-l
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### Re: Tricky limit problem for a calculus course

While I agree with the result, the method is incorrect, as applying the same logic to
$\lim_{n\to \infty} (1 + \frac 1n)^n$
gives 1, when we ought to know that the above is e.

You can't 'apply L'Hopital n times' when n is the index of the limit, since L'Hopital is only true for [imath]n\to\infty[/imath]

As a proof that it's equal to 1, I'd consider its inverse, then show it's bounded above by [imath]e^\varepsilon[/imath] (and clearly bounded below by 1). Haven't taken the time to do it but I'd wager it works.

Also you can probably show it's monotone for sufficiently large n, and if so the real number case is identical to the integer case. (Also the above sketch of a proof would work for the real number case directly)
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skeptical scientist
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### Re: Tricky limit problem for a calculus course

Qaanol wrote:Indeed this can be repeated [imath]n[/imath] times, with the last calculation being,
$L = \lim_{n\rightarrow\infty}\frac{\log(n-1)}{\log(n)} = \lim_{n\rightarrow\infty}\frac{\frac{1}{n-1}}{\frac{1}{n}} = \lim_{n\rightarrow\infty}\frac{n}{n-1} = 1$

No, no it can't. You're taking a limit as n approaches infinity, so n is a dummy variable. Using L'Hopital's rule n times is complete nonsense, since there is no fixed number of times to take it. You are just inventing a completely new sequence, which bears no resemblance to the original sequence, and so there's no valid reason to expect them to have the same limit. (Also—not that it matters—you took the derivative incorrectly.)
...and ninja'd by mike-l

I can show that the limit is 1 by manipulating inequalities. Since log is concave down, by comparing log to the value of the tangent line at n-1, we see $$\log(n-1) > \log n - \frac{1}{n-1}.$$ Thus $$\left( \frac{\log(n-1)}{\log n} \right)^n > \left( 1 - \frac{1}{(n-1)\log n} \right)^n.$$ Furthermore, for all e<1 and n>1, we have (1-e)n>1-ne. (This can be proven easily by induction.) Therefore, we have $$\left( \frac{\log(n-1)}{\log n} \right)^n > \left( 1 - \frac{1}{(n-1)\log n} \right)^n>1-\frac{n}{(n-1)\log n}>1-\frac{2}{\log n},$$ so $$1 > \left( \frac{\log(n-1)}{\log n} \right)^n > 1-\frac{2}{\log n}.$$ By the squeeze theorem, this goes to 1 as n approaches infinity.

That said, this proof is definitely more than I would expect from a first-year calc student (although doable in principle, since no techniques more advanced than calculus were used), so I'm not sure how the problem was meant to be solved.
Last edited by skeptical scientist on Tue Nov 16, 2010 8:53 pm UTC, edited 1 time in total.
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skullturf
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### Re: Tricky limit problem for a calculus course

@ mike-l: Now that I think about it, I agree with you.

L'Hopital's rule, applied once, implies that the answer (if it exists) is the same as the limit of

[log(n-1)/log(n)]^(n-1).

L'Hopital's rule applied k times implies that the answer, if it exists, is the same as the limit of

[log(n-1)/log(n)]^(n-k).

But it doesn't actually make sense to say, in this case, "Apply L'Hopital's rule n times." (edited to add: somebody beat me to it.)

@ gorcee: I don't think complex numbers come into it at all. If n is a real number greater than 2, then log(n-1) and log(n) are positive real numbers. The exponent n might be an ugly fraction or decimal, but I don't think complex numbers enter into the problem.

--------------------------

And with a blushing face, I admit that skeptical scientist is right about Qaanol not taking the derivative correctly. I didn't even notice at first.

There's a lesson here about the power of notation. If only the variable had been called x and not n, I probably would have caught the mistake.
Last edited by skullturf on Tue Nov 16, 2010 9:10 pm UTC, edited 1 time in total.

gorcee
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### Re: Tricky limit problem for a calculus course

skullturf wrote:@ mike-l: Now that I think about it, I agree with you.

L'Hopital's rule, applied once, implies that the answer (if it exists) is the same as the limit of

[log(n-1)/log(n)]^(n-1).

L'Hopital's rule applied k times implies that the answer, if it exists, is the same as the limit of

[log(n-1)/log(n)]^(n-k).

But it doesn't actually make sense to say, in this case, "Apply L'Hopital's rule n times." (edited to add: somebody beat me to it.)

@ gorcee: I don't think complex numbers come into it at all. If n is a real number greater than 2, then log(n-1) and log(n) are positive real numbers. The exponent n might be an ugly fraction or decimal, but I don't think complex numbers enter into the problem.

You said real numbers, not real positive numbers.

The inequality

$\left(\frac{\log(n-1)}{\log n}\right)^n > 1-\frac{2}{\log n}$

doesn't hold for [imath]n \le 1[/imath].

Qaanol
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### Re: Tricky limit problem for a calculus course

That’s what I get for trying to do math before breakfast.
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skullturf
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### Re: Tricky limit problem for a calculus course

gorcee wrote:
You said real numbers, not real positive numbers.

True, but I also said limit as n approaches infinity.

skeptical scientist did something similar to what I did. My estimation of the difficulty of the question was similar, too -- I think it's probably too tricky for a typical first year calculus student, even though no techniques they haven't seen are used.

I was mainly interested to see whether I was correct about the difficulty level, to see whether my solution was more or less the "natural" way to do it, and to see whether I was missing some neat and clever alternate method.

mike-l
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### Re: Tricky limit problem for a calculus course

gorcee wrote:You said real numbers, not real positive numbers.

The inequality

$\left(\frac{\log(n-1)}{\log n}\right)^n > 1-\frac{2}{\log n}$

doesn't hold for [imath]n \le 1[/imath].

That doesn't matter, since we're taking the limit as [imath]n\to\infty[/imath] we may assume n is bigger than any constant you like, in particular, n>1 is a perfectly valid assumption.

Ninja'd!
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gorcee
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### Re: Tricky limit problem for a calculus course

mike-l wrote:
gorcee wrote:You said real numbers, not real positive numbers.

The inequality

$\left(\frac{\log(n-1)}{\log n}\right)^n > 1-\frac{2}{\log n}$

doesn't hold for [imath]n \le 1[/imath].

That doesn't matter, since we're taking the limit as [imath]n\to\infty[/imath] we may assume n is bigger than any constant you like, in particular, n>1 is a perfectly valid assumption.

Ninja'd!

Durr, right. I had been thinking that the repeated derivative proof worked, but that really doesn't make much sense either.

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### Re: Tricky limit problem for a calculus course

Hmm, without actually thinking about limits, the first thing that came to mind when I saw the expression was the change of base formula, producing:

(logn(n-1))^n as n->infinity.

as n->infinity, n-1 approaches n, so we'd have logn(n)=1, leaving us with 1^infinity=1.

I haven't formally dealt with limits in ages, so I probably worded things that are technically wrong there, but I like that it at least produces the correct limit in a hand waving way at least.

skullturf
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### Re: Tricky limit problem for a calculus course

Minor quibble regarding skeptical scientist's answer: If we allow n to take on real values, then we can't prove (1-e)^n > 1-ne "by induction". However, we can still say that if k is a real number > 1 and if e is a real number between 0 and 1, then (1-e)^k > 1-ke. This can be done similarly to what skeptical scientist did with a graph and its tangent function. Specifically, consider here the graph of y = x^k which is concave up, so the tangent line at x = 1 lies below the graph. Apply this to x = 1 - e and you will get the desired result.

In my original proof, I played around with Taylor series, which is overkill if all you really care about is whether something is slightly less or slightly greater than its linear approximation.

And once again, I'd love to hear about any alternate proofs -- although I suspect we've found essentially the "right" way to do the problem.

----------------

@ Dopefish -- 1^infinity is an example of an "indeterminate form". Limit problems of this "shape" can have many different answers. A famous example is lim (1 + 1/n)^n, which happens to be e. In my lectures, I sometimes ask my students, "What do you 'expect' to happen if you raise a number like 1.000001 to the millionth power?"

You could almost say it's "unfortunate" in a way, for pedagogical reasons, when a limit problem of the form 1^infinity does happen to have 1 as the correct answer.

In my original approach to the problem, I tried to rewrite as (1 - thing1)^thing2, where thing1 was close to 0 and thing2 was large, and tried to show that the smallness of thing1 "wins" the battle against the largeness of thing2.

LaserGuy
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### Re: Tricky limit problem for a calculus course

skullturf wrote:This is not homework. In fact, I am a college-level math instructor who's composing a test today, and I'm not including this particular problem because it appears too tricky.

However, it's always possible I missed a clever shortcut.

The problem: Find the limit of ( log(n-1)/log(n) )^n as n approaches infinity.

I'm confident I know the answer, and I'm pretty sure I have a way to prove it, but I am curious if there are other ways or if I am missing a clever shortcut.

It looks like you should be able to take the logarithm of both sides and apply L'Hopital's rule (probably twice) to get rid of the logarithms. It's not terribly sexy, but it should work.

Here's a proof.
[edit2]Ignore proof and revoke my math license.
Last edited by LaserGuy on Tue Nov 16, 2010 11:51 pm UTC, edited 1 time in total.

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### Re: Tricky limit problem for a calculus course

skullturf wrote:@ Dopefish -- 1^infinity is an example of an "indeterminate form". Limit problems of this "shape" can have many different answers. A famous example is lim (1 + 1/n)^n, which happens to be e. In my lectures, I sometimes ask my students, "What do you 'expect' to happen if you raise a number like 1.000001 to the millionth power?"

Yeah, like I said, it was very hand wavy. Most of my math comes from physics courses.

I believe the whole "n-1 approaches n" line is fairly shady too. It's fine for stuff like lim n->infinity of (n-1)/(n) (and similar polynomial stuff) since that's the same as 1-1/n, but for logn(n-1) you can't just divide stuff through to make the limit obviously work out.

Come to think of it, a proper proof of the limit of logn(n+1) as n-> infinity on its own (without the surrounding ^n) might be interesting. Hmm... Could probably throw taylor series at it, but ehh...

skullturf
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### Re: Tricky limit problem for a calculus course

Dopefish wrote:
Come to think of it, a proper proof of the limit of logn(n+1) as n-> infinity on its own (without the surrounding ^n) might be interesting. Hmm... Could probably throw taylor series at it, but ehh...

That one succumbs to L'Hopital's rule very well.

I initially thought my problem could probably be done by L'Hopital's rule as well, but the algebra gets very messy, and I'm not even sure whether that approach leads to an answer in any reasonably obvious way.

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### Re: Tricky limit problem for a calculus course

Heh, I forgot about l'hopital for a minute. Yay for that magic* rule!

*= not actually magic.

++$_ Mo' Money Posts: 2370 Joined: Thu Nov 01, 2007 4:06 am UTC ### Re: Tricky limit problem for a calculus course As far as I can tell, the easiest way to do the original problem is to first take the log. Then we have $\lim_{n \to \infty} n(\log \log(n-1) - \log \log n).$ Now, what is [imath]\log \log n - \log \log (n-1)[/imath]? Basically, that is the derivative of [imath]\log \log n[/imath], which is [imath]{1 \over n\log n}[/imath]. (Technically, I guess we have to bound it above by the maximum value of the derivative between [imath]n-1[/imath] and [imath]n[/imath], so it's actually more like [imath]{1 \over (n-1)\log(n-1)}[/imath]. And $\lim_{n \to \infty}{n \over (n-1)\log(n-1)} = 0.$ Therefore, the answer to the original problem is [imath]e^0[/imath], or 1. Did I make a mistake somewhere? skullturf Posts: 556 Joined: Thu Dec 07, 2006 8:37 pm UTC Location: Chicago Contact: ### Re: Tricky limit problem for a calculus course @ ++$_

I like it!

There may be various ways to make precise the idea that log log n - log log (n-1) is "basically" the derivative of log log n. In a sense, you're comparing the slope of the secant line with the slope of the tangent line, so one could claim you're using the "same" technique, in a broad sense, as skeptical scientist and myself. But I like your way of putting it.

++\$_
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### Re: Tricky limit problem for a calculus course

Yeah, actually, it is the same technique. I managed to miss skeptical scientist's post. I just took a log first so that I could write it as a difference of two functions, which is something I find fairly intuitive, rather than as a ratio, which I don't.

To make it rigorous, I guess I'd just point out that for differentiable functions on [imath][a,b][/imath], [imath]f'(x_m)(b-a) \le f(b) - f(a) \le f'(x_M)(b-a)[/imath], where [imath]x_m[/imath] and [imath]x_M[/imath] are the points where the derivative is minimized and maximized, respectively. The proof is immediate from the mean value theorem.

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### Re: Tricky limit problem for a calculus course

Wolfram|Alpha did it "the long way," which I guess works out just fine. Whenever it encountered form [imath]1^\infty[/imath], it applied the transformation [imath]\lim_{n\to\infty}f^g = e^{\lim_{n\to\infty}g \log{f}}[/imath], whenever it encountered form 0/0 or ∞/∞, it applied L'Hospital's rule, and whenever it encountered form 0 * ∞, it substituted t = 1/n. After maybe half a dozen iterations of this, the numerator became 1 and it sorted out the (huge and ugly) denominator without much trouble.

This would be a pain in the ass to do in practice.

skullturf
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### Re: Tricky limit problem for a calculus course

@ Eebster -- That's fascinating! Thanks!

I was wondering whether or not the "straightforward" way, where you just chug through using your Calc I techniques, would work here at all. It looks like it does, but very few humans would have the patience to go through the steps.

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### Re: Tricky limit problem for a calculus course

Did anyone bother to prove that the original limit even exists? EDIT: Nevermind, the answer to that is yes.
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cameron432
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### Re: Tricky limit problem for a calculus course

I know I'm doing something wrong with this..., but:
(lim is just limit as n-->infinity)

lim ( log(n-1) / log(n) )^n )

lim ( (log(n-1)^n) / (log(n)^n) )

lim ( (nlog(n-1)) / (nlog(n)) )

lim ( (log(n)/log(1)) / (log(n)) )

lim ( (log(n)) / (log(1) * log(n))

1 / log(1)

1 / 0

limit Doesn't Exist

Again, I know my math is screwed up somehow, but I don't know. I'm in Calc 2 this semester...

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### Re: Tricky limit problem for a calculus course

cameron432 wrote:I know I'm doing something wrong with this..., but:
(lim is just limit as n-->infinity)

lim ( log(n-1) / log(n) )^n )

lim ( (log(n-1)^n) / (log(n)^n) )

lim ( (nlog(n-1)) / (nlog(n)) )

Same mistake as laserguy.
(log a)n does not equal log(an) = n log a.

agelessdrifter
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### Re: Tricky limit problem for a calculus course

Do logs not have the property that $\lim_{n\to\infty} u(n)^n = \lim_{n\to\infty} [\lim_{n\to\infty} u(n)]^n$

by which I mean, couldn't you just take the limit of the inside first? It seems fairly simple to show that $\frac{log(n)}{log(n-1)}$ goes to 1 as n goes to infinity. Then you'd just have $\lim_{n\to\infty}1^n$

I'm guessing since no one did that, it's because logs don't have that property.

jestingrabbit
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### Re: Tricky limit problem for a calculus course

In general, $\lim_{n\to\infty} u(n)^n \neq \lim_{n\to\infty} [\lim_{m\to\infty} u(m)]^n.$

For instance, $\lim_{n\to\infty} \left(1 + \frac{1}{n} \right)^n = e \neq 1 = \lim_{n\to\infty} \left(\lim_{m\to\infty} 1 + \frac{1}{m} \right)^n.$

No one did that because its not correct.
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### Re: Tricky limit problem for a calculus course

Ya know, I sat here and tried to think of a counter-example for several minutes before I posted. I feel pretty ridiculous for not considering e, which is about as obvious as it gets

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### Re: Tricky limit problem for a calculus course

cameron432 wrote:
lim ( (nlog(n-1)) / (nlog(n)) )

lim ( (log(n)/log(1)) / (log(n)) )

Also, log(a-b) is not the same as log(a)/log(b), if that's one thing you were trying to do there.