/sci/ - Science & Math

[Anonymous No.4325954]

Is this valid?

<span class="math"> (\frac {3}{n})^n = (\frac {3}{n})^n [/spoiler]

<span class="math"> (\frac {3}{n})^n = 3^n * (\frac {1}{n})^n[/spoiler]

<span class="math"> \lim_{n \to \infty} (3^n * (\frac {1}{n})^n) = (\lim_{n \to \infty} (3^n)) * (\lim_{n \to \infty} (\frac {1}{n}))^n[/spoiler]

<span class="math">\to \lim_{n \to \infty} (3^n) * (0)^n[/spoiler]

<span class="math"> = 0 [/spoiler]

I'm not sure about whether it is valid to do the third to last step.

[Anonymous No.4325957]

>>4325954
Lol. I always seem to make stupid mistakes.

The first line of course is to be read as <span class="math"> \frac {3^n}{n^n} = (\frac {3}{n})^n[/spoiler]

[Anonymous No.4325971]

You need to use L'Hospital's rule

[Anonymous No.4325972]

It isn't, limits aren't distributive.

[Anonymous No.4325982]

>>4325972
Multiplicative you mean I think. Besides he didn't even do n^n properly.

[Anonymous No.4326019]

>>4325972
Uh.. I thought that <span class="math"> \lim_{x \to z} (f(x) * g(x)) = \lim {x \to z} f(x) * \lim_{x \to z} g(x) [/spoiler]

[Anonymous No.4326024]

>>4326019
that's fine if both limits exist. But 3^n as n goes to infinity certainly doesn't exist.

[Anonymous No.4326025]

>>4326019

Only if the limit exists

[Bruised No.4326026]

>Algebra of limits on two sequences that don't converge.

Use L'Hôpital's rule.

[Anonymous No.4326027]

>>4325972

[Anonymous No.4326031]

>>4326025
D'oh. My bad, overlooked that.

[Anonymous No.4326034]

>>4326027
Meant that they're not always multiplicative, I freely admit to fucking up. Just woke up, still in bed blegh.

[Anonymous No.4326038]

Only the two of us seem to know L'Hospital's rule

[Anonymous No.4326044]

>>4326038

LOL. L'Hopital's rule is for pussies. You can evaluate it without using L'Hopital.

[Anonymous No.4326048]

>>4326044
cept you can't.

[Anonymous No.4326046]

>>4326038

He didn't ask how to do it, so I think most people wouldn't feel compelled to interject with an alternate solution path. You're not amazing for having taken calc 1.

[Bruised No.4326052]

ITT

[Anonymous No.4326054]

>>4326046
Nor are you for pointing it out. He would likely be stuck if I simply said no you cannot do the third step. He hasn't already though of L'Hospital's rule so it's likely it didn't occur to him that this is a situation that required it. He would end up back here very soon, asking how to go about solving it again.

[Anonymous No.4326055]

>>4326048

Not the guy that you're responding to, and I doubt this is what he meant (people who boast about avoiding the most appropriate mathematic techniques usually aren't very good at math in my experience)- there's a ton of alternate methods of evaluating limits that probably didn't get covered in your undergrad calculus courses, many of which are applicable to this situation.

[Anonymous No.4326058]

>>4326054

Right, and it's fine that you did say something- I never suggested that you shouldn't have. I just said that most people wouldn't feel the need to because it wasn't asked for, so you shouldn't take a lack of silence on that grounds as an indication that you hold some special knowledge that can only be obtained by taking introductory calculus for a few weeks. Also, there's not much reason to suggest it after it's already been suggested.

[Anonymous No.4326064]

>>4326058
Fair enough, but I took the lack of silence as people possibly confusing OP more.

[Anonymous No.4326080]

>>4326064

Makes total sense, I didn't think to try factoring polynomials until >n people told me.

[Anonymous No.4326081]

>>4326064
OP here.

I'm not exactly confused as to why you can (or have to?) use L'hospital's rule, since you use it when a term results in an undefined state, e.g. <span class="math"> \frac {0}{0} or \frac {\infty}{\infty} [/spoiler], which is the case because \frac <span class="math">{3^n}{n^n}[/spoiler] is such a case, if I'm not mistaken.

Yet, we've only mentioned the rule, and my professor said that there are other ways to solve it, and that he'd rather want us to figure them out as well.

[Anonymous No.4326093]

>>4326081
You have <span class="math">\frac{3^n}{n^n}[/spoiler] which is the infinity / infinity form of your <span class="math">3^n (\frac{1}{n})^n[/spoiler]

[Anonymous No.4326100]

>>4326093
Yup. But since my "distributive limits" argument was voided by someone pointing out that it's only true when both factors have a limit, I'm unsure about what to do now.

[Anonymous No.4326101]

>>4326100

Take the n root of top and bottom

[Anonymous No.4326102]

>>4326100
Which is why I pointed out that L'Hospital's rule applies here. Was hoping it'd get you thinking to put it into the form I did above instead after.

[Anonymous No.4326106]

>>4326101
This works too iirc

[Anonymous No.4326112]

>>4326101
Somehow I feel really stupid for not having noticed that.

[Anonymous No.4326114]

>>4326106
No, because you'd have a limit that doesn't exist on the other side of the equation. L'Hospital's rule again.

[Anonymous No.4326118]

>>4326114

the nth root is 1/n, which goes to zero. c^(1/n) =1

[Anonymous No.4326126]

>>4326118
Which of course is false, because application of L'hospital's rule -something appropriate for this situation- shows that that is not the solution.

[Anonymous No.4326135]

>>4326118

>>4326126

For instance,
lets say you're taking the limit as n goes to infinity of 1/2^n
if you applied the n'th root thing, you would get the limit is 1 again, but just doing it straight out, it wouldn't work because the solution is clearly 0

[Anonymous No.4326138]

Funnily enough, this all started out as the idea of checking whether the series <span class="math"> \sum \limits_{n=1}^{\infty} \frac {3^n}{n^n} [/spoiler] converges or not, simply checking if its sequence is a null sequence. But THAT turned out to be far more difficult than just using convergence criteria

[Anonymous No.4326141]

>>4326093
<span class="math">e^{nlog(\frac{3}{n})}[/spoiler]
inside exponential is neither <span class="math">\frac{0}{0}[/spoiler] nor <span class="math">\frac{\infty}{\infty}[/spoiler]
lol does Lhospital rule apply anymore ?

[Anonymous No.4327287]

>>4326141
Yes, because that is unnecessary. The problem is already given to you in infinity/infinity form more or less converting it to a 0* infinity form is just.. dumb.

[Anonymous No.4327292]

You can only apply the calculus of limits theorem in the way you did in third line if all of the separated terms tend to a limit. 3^n tends to infinity so it is invalid.

[Anonymous No.4327309]

>>4326138
Pretty sure it's just zero bro

[Anonymous No.4327331]

That's not the problem, it's because 0*inf is not defined.

[Anonymous No.4327383]

. you cant say lim f*g=limf*limg except if all limits exists. the result sure is 0. intuitivily, well its a product of more and more smaller terms. with a cleaner reasonnement just say that the lim of the log ie (n*(ln3-ln n) tend toward -inf

[Anonymous No.4327404]

>invoking some limits before others, then getting a wrong answer

Wow.

[Anonymous No.4327890]

>>4327287
is it 0* infinity in the exponential though ?
>highschool

[Anonymous No.4327898]

>>4326141
>>4326141
>>4326141
lol and the dumbasses think Lhospital is the only way to solve this limit

[Anonymous No.4327905]

>>4327287
>>4327287

[Anonymous No.4327922]

>>4327898
Except that way doesn't work.
You're free to try again.

[Anonymous No.4327956]

>>4327922
limit of e^u when u approach negative infinity

[Anonymous No.4327961]

>>4326141

>>4327922
it works exactly as intended
hope i dont have to spell out how to proceed from there

[Anonymous No.4328033]

>>4327961
If you would, please

>>4327956
except in this case you have u = infinity times negative infinity, which itself is an indeterminate form and you're in trouble once again.

[Anonymous No.4328079]

>>4328033
>doesnt know the difference between undefined and indeterminate

here u go :
<span class="math">\lim_{n \to \infty} e^{n\ln(\frac{3}{n})} = \lim_{n \to \infty} \frac{1}{e^{n\ln(\frac{n}{3})} }[/spoiler]
the expression
<span class="math">a(n) = nlog(\frac{n}{3})[/spoiler] grows unbounded as n grows unbounded
proof: for every n choose n' > n then a(n') > a(n) etc
therefore the original limit is 0

if you don't get that you should retake highschool

[Anonymous No.4328087]

>>4325954
No, you're an idiot.

[Anonymous No.4328095]

>>4328079
Yes that does work

And the abrasive attitude is absolutely unnecessary.

[Anonymous No.4328106]

>>4328095
oh i'm sorry then

[Anonymous No.4328148]

>>4328106
You'll pardon me because Im not use to messing with limits. I know of that e^ln trick, but i didn't know to use it here, though i tried applying it to some other unrelated problem few days ago.

But yeah thanks. Its good to reinforce knowedge foundations.

[Anonymous No.4329476]

Use the squeeze theorem with

<span class="math">f(x)=\frac{3^x}{x^x},g(x)=\frac{1}{3^x},h(x)=\frac{1}{x^x}[/spoiler]