limits, yes another limits thread.

MathematicalPhysicist

i need to compute lim (a^n-b^n)^(1/n) when a>b>0.
lim ((3n)!/((2^3n)n!(2n!)))^(1/n) where i need to use the lemma that:
if an>0 for every n, and lim(x_n+1/x_n)=L then lim x_n^(1/n)=L, how to use it here?

for the first i used practically everything i know, the formual for a^n-b^n, and the fact that 0<a^n-b^n<a^n and lots more algebraic techniques, apparently not everything.

your help is appreciated.

MathematicalPhysicist

another thing, seemingly unrealted.
i need to show that for 0<x<2pi x>=sinx.
i tried to dissect it into parts, i.e:
x>1 and for x<1, for x>1 it's obvious, my problem is with x<1, then how to solve it?

arildno

It is relatively simple to show that there IS a limit, and that it lies between a and b, however any more precise than that, I don't know.

Use the following identity in a clever manner:
[tex]a^{n}-b^{n}=(a-b)\sum_{i=0}^{n-1}a^{n-1-i}b^{i}[/tex]

HallsofIvy

The last one, to show that [itex]x\ge sin x[/itex] for all [itex]0\le x\le 2\pi[/itex] is easy. What is the derivative of f(x)= x- sin(x)? For what values of x is that positive? Since sin(0)= 0, what does that tell you?

MathematicalPhysicist

oh, come on halls, you don't know a proof which doesnt employ derivatives, sure it's easy with derivative but i want to show it without.

arildno

Well, what's wrong with a geometrical argument on the unit circle, then?

MathematicalPhysicist

as i said i tried this equality, i got something like this: (a-b)a^n>=(a-b)(a^(n-1)+a^(n-2)b+....+b^n-2a+b^(n-1))>=(a-b)(b^(n-1)+...+b^(n-1))=(a-b)b^n but as you said that's easy to show, the problem is to what does it converge?
any other tips?

btw, i also need some help in the other limit.

as always your help is appreciated.
(my custom mantra (-: ).

MathematicalPhysicist

i dont follow you, how to use the unit circle here?
i mean the identity which seems to be good here is sin^2x+cos^2x=1
sin^2x=1-cos^2x
sinx/x=sqrt(1-cos^2x)/x<=1 when 0<sinx<=1
cause 1-x^2<=cos^2x<=1
is this correct?

arildno

Remember that in general, we have NO infallible technique that enables us to "calculate" a limit.
For practical purposes (say when solving a non-linear system numerically), we use the Cauchy criterion to say we have "essentially" reached the limit.
But formally, of course, this is bogus.


What we DO have, is first and foremost a PROPERTY that the limit must have. It is by no means guaranteed by this that we actually manage to FIND that number.

MathematicalPhysicist

so to sum up, you don't know how to calculate the limit, correct?
if it's any good, i got a hint to use the sandwich lemma.

arildno

Correct!
I already said that in my first post.

arildno

No. What length does "sin(x)" represent on the unit circle, and where can you find the length "x" on the same circle?
Compare those lengths!

MathematicalPhysicist

ah, i see your point. x is the length of the arc and sinx is the length of the y coordinate on the coordiantion system.

arildno

Of course, the geometrical argument rests also upon the assumption that the straight line is the shortest distance between two points..

MathematicalPhysicist

about the limit with a and b, i think it converges to a, the problem is how to prove it, obviously [tex](a^n-b^n)^{\frac{1}{n}}<a[/tex]
one way to show that it converges to a is by the sandwich lemma, the problem is to find the suitable lower bound, i thought perhaps a^n/n can be a good candidate, but im not sure this is correct, can someone help?

matt grime

It's always easier to get rid of two things that depend on n, and replace it with one.

What can you say about the convergence, or otherwise, of

[tex] (1-r^n)^{1/n}[/tex] where 0<r<1?

It is quite straight forward just using the most naive bounds there are for (1-r^n) (assuming we don't think 0 is a naive lower bound, I mean).

MathematicalPhysicist

it converges to 1, is it not?
anyway, i need to find a suitable lower bound for the limit.