Evaluating Limits: Understanding Why lim n->inf sin(n)^(1/n) = 1

[GreenPrint]

Why is it that lim n->inf sin(n)^(1/n) = 1? I can't seem to figure it out why, thanks for any help that you can provide.

 

[stallionx]

Why is it that lim n->inf sin(n)^(1/n) = 1? I can't seem to figure it out why, thanks for any help that you can provide.

What kind of values would Sin(n) take where whatever (n) is ?

 

[GreenPrint]

All values from -1 to 1, including these values, and then you would raise all those values to ^(1/n), were n is all values from 1 to infinity including 1, still unsure why though

 

[GreenPrint]

I checked with wolfram alpha
http://www.wolframalpha.com/input/?i=lim+n-%3Einf+sin%28n%29^%281%2Fn%29
it claims that lim n->inf sin(n)^(1/n)=1

 

[lanedance]

its an interesting one

now do you need to prove it or just convince yourself qualitatively?

first what is the limit of
$$\lim_{n \to \infty}(-1)^{1/n}$$

in fact for any number a, not zero, what is
$$\Lim_{n \to \infty}(a)^{1/n}$$

another thing that is instructive to do might be to consider the magnitude in the limit

The problems with sin will be around the zero, or n = k.pi, for integers k. For large n the exponent will "squeeze" these points of divergence to an infinitesimal region

 

[dynamicsolo]

its an interesting one

now do you need to prove it or just convince yourself qualitatively?

first what is the limit of
$$\lim_{n \to \infty}(-1)^{1/n}$$

in fact for any number a, not zero, what is
$$\Lim_{n \to \infty}(a)^{1/n}$$

another thing that is instructive to do might be to consider the magnitude in the limit

The problems with sin will be around the zero, or n = k.pi, for integers k. For large n the exponent will "squeeze" these points of divergence to an infinitesimal region

It is correct to say that you will need the "Squeeze Theorem" in some way, because L'Hopital's Rule will be useless since $\lim_{n \rightarrow \infty} \sin(n)$ does not exist.

However, you will need to avoid using the problematic function $a^{1/n}$ , with a < 0 : it has a countably infinite number of undefined points, those being where n is even. (I notice, in fact, that Grapher won't even attempt y = (-1)^1/x .) The proof will need to be drawn a bit more carefully...

 

[GreenPrint]

I originally thought that as well

lim n->inf a^(1/n) = 1

so

lim n->inf sin(n)^(1/n) = 1

but I wasn't so sure about that because lim n->inf sin(n) is undefined but would contain some value between -1<= sin(inf) <=1, at least I think

and you interestingly enough for some reason (-1)^(1/inf) is -1 and not positive one, not really sure why, (-1)^(1/inf) should be equal to (-1)^0 and should be equal to one 1 but it's not and actually negative one, not really sure why it's -1 and not 1, 0^(1/inf)=0^0 is undefined as well.

 

[Dick]

If you mean lim n->inf sin(n)^(1/n) where n is a real number then to say it's 1 is silly. If n=k*pi, then sin(k*pi)=0 and 0^(1/n)=0. That's not near 1. If you mean n to be an integer then it's not so obviously wrong because k*pi is never an integer, but it can be REALLY CLOSE to an integer.

 

[Dick]

I originally thought that as well

lim n->inf a^(1/n) = 1

so

lim n->inf sin(n)^(1/n) = 1

but I wasn't so sure about that because lim n->inf sin(n) is undefined but would contain some value between -1<= sin(inf) <=1, at least I think

and you interestingly enough for some reason (-1)^(1/inf) is -1 and not positive one, not really sure why, (-1)^(1/inf) should be equal to (-1)^0 and should be equal to one 1 but it's not and actually negative one, not really sure why it's -1 and not 1, 0^(1/inf)=0^0 is undefined as well.

Why would you think lim n->inf (-1)^(1/n)=(-1)?? First of all you have to DEFINE what (-1)^(1/n) means since there are n different nth roots. By making a sufficiently bizarre choice I think you can make the limit come out to any complex number z such that |z|=1. The usual principal root choice would make it +1.

 

[dynamicsolo]

Actually, having thought about this general term a bit more and making a graph of it, I think the correct answer is that this limit does not exist. The function $f(x) = (\sin x)^{1/x}$ is undefined exactly half the time , which means that it has no meaningful value at half of the integers n .

In the intervals where this function is defined, it ranges from 0 to 1 , with the associated curve coming to resemble increasingly a rectangle of unit height. The function is still only equal to 1 at odd multiples of pi/2 and becomes closer and closer to that value over a broader section of the defined intervals, but I don't think one can consider that to define a limit. (Try doing a classical "epsilon-N" proof on this critter...)

I will again point to the deficiencies of computer algebra/calculus systems in the fact that Wolfram Alpha actually claims a limit at infinity for this function. You really shouldn't take the word of a "chunk of code" too seriously...

 

[lanedance]

I think if you make some assumptions you can start to justify a limit,

so following on from Dick and trying to deal with points on multivalued functions, if you put in the following caveats:
- consider finding the limit of the series: sin(n)^(1/n) for n = 1,2,3,...
- as exponentiating to a fraction is in general a multivalued operation, choose the principal root

the fact any integer n is never equal to k.pi removes the issues with zeroes of sin(n)
choosing the principal root means removing the issues with a multi-valued function

in the previous post I was not laying out a blue-print for a comprehensive proof, but asking the OP what they wanted to achieve, and hopefully giving some insights that should help find them to a way to approach the problem

so for the limit
$$\lim_{n \to \infty }(-1)^\frac{1}{n}$$

why not consider $-1 = e^{i \pi}$, whilst acknowledging this is "choosing" a branch when the operation is multi-valued , which I'll return to later.
$$\lim_{n\to \infty }(-1)^\frac{1}{n} <br /> =\lim_{n \to \infty }(e^{i \pi})^\frac{1}{n} <br /> =\lim_{n \to \infty }e^{i\frac{\pi}{n}}$$

in this form the limit becomes clear

now in terms of the multivalued function, for the case $-1^{\frac{1}{n}}$ it is true we could choose $-1 = e^{i (2k-1) \pi}$ for k = 1,2,3... which gives $$\lim_{n\to \infty }(-1)^\frac{1}{n} <br /> =\lim_{n \to \infty }(e^{i (2k-1) \pi})^\frac{1}{n} <br /> =\lim_{n \to \infty }e^{i\frac{(2k-1) \pi}{n}} = 1, \ for \ \ finite \ \ k$$

It only really makes sense choosing k such that 0<(2k-1)<n, and if the series starts from 1, the natural choice is probably k=1. And for any choice of finite k the limit still tends to 1.

In fact if you think of it as a function $f(x) = (-1)^{\frac{1}{x}}$ then changing this k value would in fact introduce discontinuities into what is a continuous function with f(1) = -1. That forces us to set a finite k value, in effect choosing a single-valued branch of the function. All the paths have f(1) = -1, and appear to tend to 1 for large n.

If you consider plotting the function in the complex plane along the x axis, in a pseudo-3D sense, then I think they are are different "stretched spirals" in the complex plane with starting at f(1) = -1 converging on 1 for large x, with |f(x)|=1. The number of half revolutions of the x-axis is (2k-1).

All that said its been a while since I've played with multi-valued functions, so feel free to look for any mis-assumptions/inconsistencies.

Now if you consider that n is an integer, then we know sin(n) is non-zero and for any n we for some real a, with 0<a<1 we can can write:

$$sin(n) = a$$
or
$$sin(n) = a(-1)$$

 

[JHamm]

As n approaches infinity 1/n approaches 0 and anything to the 0 is equal to 1

 

[dynamicsolo]

As n approaches infinity 1/n approaches 0 and anything to the 0 is equal to 1

Well, not just anything: if the base "grows as fast" as the exponent is shrinking, you can get a constant that isn't 1 . Example: $lim_{x \rightarrow \infty} (1 + e^{x})^{1/x} = e$ . If the base "grows faster", the function can run off to infinity -- for instance, $lim_{x \rightarrow \infty} (1 + e^{x^{2}})^{1/x} = +\infty$.

 

[JHamm]

Ah, thankyou :) Sorry about that GreenPrint