## limits, yes another limits thread.

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.

 another thing, seemingly unrealted. i need to show that for 0=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?
 Recognitions: Gold Member Homework Help Science Advisor 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: $$a^{n}-b^{n}=(a-b)\sum_{i=0}^{n-1}a^{n-1-i}b^{i}$$

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## limits, yes another limits thread.

The last one, to show that $x\ge sin x$ for all $0\le x\le 2\pi$ 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?
 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.
 Recognitions: Gold Member Homework Help Science Advisor Well, what's wrong with a geometrical argument on the unit circle, then?

 Quote by 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: $$a^{n}-b^{n}=(a-b)\sum_{i=0}^{n-1}a^{n-1-i}b^{i}$$
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 (-: ).

 Quote by arildno Well, what's wrong with a geometrical argument on the unit circle, then?
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?
 Recognitions: Gold Member Homework Help Science Advisor 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.
 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.

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 Quote by loop quantum gravity so to sum up, you don't know how to calculate the limit, correct? i
Correct!
I already said that in my first post.

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 about the limit with a and b, i think it converges to a, the problem is how to prove it, obviously $$(a^n-b^n)^{\frac{1}{n}}  Recognitions: Homework Help Science Advisor It's always easier to get rid of two things that depend on n, and replace it with one. What can you say about the convergence, or otherwise, of [tex] (1-r^n)^{1/n}$$ where 0