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The hard limits thread

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Reply 40
Avatar for TeeEm
TeeEm
19
Original post by Gregorius
Deserves a rep just for the latex skills...


Definitely.
Reply 41
Avatar for atsruser
atsruser
OP
11
Original post by Kummer
Let x=1+1yx = 1+\frac{1}{y} then
Unparseable latex formula:

\displaystyle \ell = \lim_{y \to \infty} \left(1+\frac{1}{y}\right)^{ y \left(1+\frac{1}{y}\right)^{-y} } = \lim_{y \to \infty} \left{ \left(1+\frac{1}{y}\right)^y \right}^{{\dfrac{1}{ \displaystyle \lim_{y \to \infty} \left(1+\frac{1}{y} \right)^{y}} }} = e^{1/e}.



Right. Not too hard, but I thought it was nice. For added bonus points, quote a theorem that explains why that works.

BTW, that was my attempt at making something interesting and original out of the boring e-as-a-limit thing. Marks out of ten?
Reply 42
Avatar for Gome44
Gome44
10
Original post by atsruser
Right. Not too hard, but I thought it was nice. For added bonus points, quote a theorem that explains why that works.

BTW, that was my attempt at making something interesting and original out of the boring e-as-a-limit thing. Marks out of ten?


(e^(1/e))/10
Reply 43
Avatar for atsruser
atsruser
OP
11
Original post by Gome44
(e^(1/e))/10


:smile:

If I'd known that was how the marking scheme was going to work, I'd have made the upper denominator x−1x-1 as well.
Reply 44
Avatar for atsruser
atsruser
OP
11
Find lim⁡x→(2n+1)π2cosectan⁡2xx\displaystyle \lim_{x \to \frac{(2n+1)\pi}{2}} \text{cosec}^{\tan^2 x} x
Reply 45
Avatar for atsruser
atsruser
OP
11
Find lim⁡n→∞∑k=1n14n2−k2\displaystyle \lim_{n \to \infty} \sum_{k=1}^n \frac{1}{\sqrt{4n^2-k^2}}
Find

lim⁡n→∞∑m=onm2arctan⁡(m)∑m=0n(m+n)2\displaystyle \lim_{n \rightarrow \infty} \dfrac{\sum_{m=o}^{n} m^2 \arctan(m)}{\sum_{m=0}^{n} (m+n)^2}
(edited 8 years ago)
Reply 47
Avatar for atsruser
atsruser
OP
11
Find
Unparseable latex formula:

\displaystyle \lim_{x \to \infty} \Big( \frac{\log(e \log x)}{\log(\log x)} \Big )^\log(\log^e(x))

Original post by Indeterminate
Find

lim⁡n→∞∑m=0nm2arctan⁡(m)∑m=0n(m+n)2\displaystyle \lim_{n \rightarrow \infty} \dfrac{\sum_{m=0}^{n} m^2 \arctan(m)}{\sum_{m=0}^{n} (m+n)^2}


It looks as though people are not so keen on limits as they are on integrals :lol:

Anyway, this problem can be tackled quite easily. Observe that

lim⁡n→∞∑m=0nm2arctan⁡(m)∑m=0n(m+n)2=lim⁡n→∞1n∑m=0n(mn)2arctan⁡(m)1n∑m=0n(1+mn)2\displaystyle \lim_{n \rightarrow \infty} \dfrac{\sum_{m=0}^{n} m^2 \arctan(m)}{\sum_{m=0}^{n} (m+n)^2} = \lim_{n \rightarrow \infty} \dfrac{\frac{1}{n} \sum_{m=0}^{n} \left(\dfrac{m}{n}\right)^2 \arctan(m)}{\frac{1}{n} \sum_{m=0}^{n} \left(1+\dfrac{m}{n} \right)^2}

=lim⁡n→∞1n∑m=0n(mn)2arctan⁡(m)lim⁡n→∞1n∑m=0n(1+mn)2=\displaystyle \dfrac{\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{m=0}^{n} \left(\dfrac{m}{n}\right)^2 \arctan(m)}{\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{m=0}^{n} \left(1+\dfrac{m}{n} \right)^2}

=π2(∫01x2 dx)(∫01(1+x)2 dx)−1 = \displaystyle \dfrac{\pi}{2} \left(\int_{0}^{1} x^2 \ dx\right)\left(\int_{0}^{1} (1+x)^2 \ dx\right)^{-1}

=π14= \dfrac{\pi}{14}
(edited 8 years ago)
Show that lim⁡n→∞ sin⁡n\displaystyle \lim_{n\to\infty}\,\sin n does not exist.
Original post by Lord of the Flies
Show that lim⁡n→∞ sin⁡n\displaystyle \lim_{n\to\infty}\,\sin n does not exist.


Shame it is not enough to mention that it oscillates between -1 and 1.

Assume the others will find a solution using the epsilon-delta definition of a limit.
(edited 8 years ago)
Reply 51
Avatar for atsruser
atsruser
OP
11
Original post by Kvothe the arcane
Shame it is not enough to mention that it oscillates between -1 and 1.


It is enough. If a sequence converges, then all subsequences of it converge, and converge to the same limit.

The contrapositive of that statement says that if 2 subsequences converge to different limits, then the sequence does not converge.

But this is LoTF, so he's probably got something tricksy up his sleeve ...
Reply 52
Avatar for Zacken
Zacken
22
Original post by atsruser
It is enough. If a sequence converges, then all subsequences of it converge, and converge to the same limit.

The contrapositive of that statement says that if 2 subsequences converge to different limits, then the sequence does not converge.

But this is LoTF, so he's probably got something tricksy up his sleeve ...


There's a nice proof of it I vaguely remember that uses (xn,yn)=(cos⁡n,sin⁡n)(x_n, y_n) = (\cos n, \sin n) and then applies the additional formula and takes the limit to get a proof by contradiction. If you're interested, I could dig it up.
Original post by atsruser
It is enough. If a sequence converges, then all subsequences of it converge, and converge to the same limit.


No red herrings/weird detours with this one, the approach you mention is exactly what I had in mind. :biggrin: Finding those subsequences isn't dead easy though.
Reply 54
Avatar for atsruser
atsruser
OP
11
Original post by Lord of the Flies
No red herrings/weird detours with this one, the approach you mention is exactly what I had in mind. :biggrin: Finding those subsequences isn't dead easy though.


No, I see what you mean. I guess this is a matter of finding subsequences where n→mπ/2,(2m+1)π/2n \to m\pi/2, (2m+1)\pi/2 for some clever choice of mm, or something like that?
Original post by atsruser
...


Prove that for any irrational α∈R,  n\alpha\in\mathbb{R},\;n and αm\alpha m can get arbitrarily close. In other words dist(Z∗,αZ∗)=0\text{dist}(\mathbb{Z}^*, \alpha\mathbb{Z}^*)=0, which is a nice result.

(note this implies cl {sin⁡n  ∣  n∈N}=[−1,1]:  \text{cl}\,\{\sin n\;|\;n\in\mathbb{N}\}=[-1,1]:\; so sin⁡n\sin n has uncountably many limits points)

Though the statement above is stronger, the proof is the same as that required for particular choice(s) of α\alpha i.e. those in your post - so might as well!
(edited 8 years ago)
Are there any non-integers x>1x>1 that satisfy lim⁡n→∞{xn}=0?\displaystyle\lim_{n\to\infty} \{x^n\}=0?
(edited 8 years ago)
Reply 57
Avatar for atsruser
atsruser
OP
11
Original post by Lord of the Flies
Prove that for any irrational α∈R,  n\alpha\in\mathbb{R},\;n and αm\alpha m can get arbitrarily close.


I think I shall have to pull a "TeeEm" here, and say, "get one of the purists to do it", as it's been far too long since I've had to prove something like that. In fact, at times like this I praise the Lord for including theoretical physics in maths curricula. But if you're happy to accept a rather nice proof by incredulity, I'd say that we seem to be requiring that we can find an arbitrarily good rational approximation to a real, and I'd be very surprised indeed if someone hasn't already show that that is true. I mean, it stands to reason, doesn't it?

In other words dist(Z,αZ)=0\text{dist}(\mathbb{Z}, \alpha\mathbb{Z})=0, which is a nice result.

Not familiar with your notation here: what is dist?

(note this implies cl {sin⁡n  ∣  n∈N}=[−1,1]:  \text{cl}\,\{\sin n\;|\;n\in\mathbb{N}\}=[-1,1]:\; so sin⁡n\sin n has uncountably many limits points)

I find that quite surprising, I must admit, but I guess obvious given your statement above.
12a\frac{1}{2 \sqrt{a}}
(edited 8 years ago)
Original post by atsruser
A level students only

Find lim⁡x→0x+a−ax\displaystyle \lim_{x \rightarrow 0} \frac{\sqrt{x+a}-\sqrt{a}}{x} with a>0a > 0 without appealing to L'Hopital's Theorem.


12a\displaystyle\frac{1}{2 \sqrt{a}}
(edited 8 years ago)

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