• MathematicalPhysicist
In summary, the conversation revolves around calculating limits using various techniques such as algebraic methods, geometrical arguments, and the Cauchy criterion. The participants discuss the use of an identity involving a^n-b^n and the sandwich lemma to calculate a limit and also consider the convergence of (1-r^n)^(1/n) where 0<r<1. There is also a mention of using the unit circle and the assumption that the shortest distance between two points is a straight line in geometrical arguments. The conversation ends with a suggestion to think about the relation between r, a, and b in order to find a suitable lower bound for the limit.
MathematicalPhysicist
Gold Member
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<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?

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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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 doesn't employ derivatives, sure it's easy with derivative but i want to show it without.

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

arildno said:
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 (-: ).

arildno said:
Well, what's wrong with a geometrical argument on the unit circle, then?
i don't 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?

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.

loop quantum gravity said:
so to sum up, you don't know how to calculate the limit, correct?
i

Correct!
I already said that in my first post.

loop quantum gravity said:
i don't 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?

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!

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.

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

about the limit with a and b, i think it converges to a, the problem is how to prove it, obviously $$(a^n-b^n)^{\frac{1}{n}}<a$$
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 I am not sure this is correct, can someone help?

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

$$(1-r^n)^{1/n}$$ 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).

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

matt grime said:
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

$$(1-r^n)^{1/n}$$ 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).
obviously it's smaller than 1, but from what it's bigger?
it's bigger than 1-r^n cause (1-r^n)^n<=1-r^n and 1-r^n<1.
will this also work fro number besides 1 and r<1, i mean:
0<b<a (a^n-b^n)^(1/n)<a, but is it: a^n-b^n<(a^n-b^n)^(1/n)
in order for to be true, a^n-b^n should be smaller than 1.

Think about it a little bit longer, please... For instance, have you figured out what relation I might intend r to have with a and b? How did I get from 0<b<a to 0<r<1?Why have you not used the fact that r^n<r , for instance? You don't appear to have thought the suggestion through at all.

well you replaced r with b and 1 with, and i did you use the fact that r^n<r when i said that (1-r^n)^n<1-r^n obviously r^n<r<1 and thus also what i wrote is correct.
anyway, i think i understand.
r^n<r
then 1-r^n>1-r so we have a lower bound and an upper.
but my question still stands, how can i use it here?
i mean a>b>0 so a^n>a^n-b^n but i cannot say anything from lower besides 0, cause I am not given that b<1.

perhaps you want to dissect this problem into situations where 0<b<a<1
0<b<1<a 1<b<a, i thought about this and obviously this is the way to go, but i thought perhaps there's a starightforward approahc which doesn't employ splitting into three situations.

No, I do not want to restrict a and b at all. My answer proves the general case for all a and b - you really can't see what to do to (a^n-b^n)^{1/n} to get something like (1-r^n)^{1/n}? I don't believe that.

geez louis, ok tell me if this is correct (sorry for not seeing it earlier):
(a^n-b^n)^(1/n)=a(1-(b/a)^n)^(1/n) b/a<1.

****,****
bloody ****. how the bloody hell this little simple thing has alluded me?!

thanks matt.

i have a tendency to make a simple problem harder than it is.

quick question:
lim
x--> -1 f(x) x + 3/|x - 2|
my answers were 2/3, and -2/3, so as there is no unique value the limit does not exist. am i right or wrong? please help:)

J-Girl said:
quick question:
lim
x--> -1 f(x) x + 3/|x - 2|
my answers were 2/3, and -2/3, so as there is no unique value the limit does not exist. am i right or wrong? please help:)

Wrong.

Limx→-1|x - 2 | = 3, for both cases: x approaches -1 from the right and x approaches -1 from the left. In either case, x approaches -1.

I deleted my reply because I noticed that this thread has been around since 2006

What is a limit in mathematics?

A limit in mathematics is a fundamental concept that describes the behavior of a function as its input approaches a certain value. It is often used to determine the value of a function at a point where it is not defined or to analyze the behavior of a function near a particular point.

How do you calculate a limit?

To calculate a limit, you can use various methods such as substitution, factoring, and L'Hospital's rule. The general process involves plugging in values closer and closer to the desired input and observing the resulting outputs to determine the limit value.

What is the difference between a one-sided limit and a two-sided limit?

A one-sided limit only considers the behavior of a function as it approaches a particular point from one side, either the left or the right. A two-sided limit, on the other hand, takes into account the behavior of a function from both the left and the right side of the point.

Why are limits important in calculus?

Limits are essential in calculus because they allow us to define and analyze important concepts such as continuity, differentiability, and convergence. They also provide a way to solve complex problems that involve functions with discontinuities or infinite behavior.

What are some real-life applications of limits?

Limits have various applications in fields such as physics, engineering, and economics. For example, they are used to calculate instantaneous rates of change in motion problems, to determine the stability of structures, and to analyze supply and demand in economics.

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