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# Sometime ago this problem was asked here and one poster ran a computer prgram which carried out 10000 multipl?

Sometime ago this problem was asked here and one poster ran a computer program which carried out 10000 multiplications on randomly selected values of x which returned very small values for each of the products. This suggested to the poster that the infinite product was zero, and she was led to the argument that since the absolute value of each term was at most 1, the limit of the product was zero. Her answer convinced the asker, who chose it as best answer. I gave a proof which yielded the answer sinx/x, and in spite of going over my steps several times, can't find the error. I should be very interested if you guys out there could tell me where I went wrong. Here is the problem and my proof.

Find Limcosx/2cosx/4cosx/8......cosx/2^n..as n→∞.

Multiply top and bottom by 2^nsinx/2sinx/4...sinx/2^n..

and rearrange as follows:

{(2sinx/2)(cosx/2)}{2sinx/4)(cosx/4}...{(2sinx/2^n)

(cosx/2^n)}.../{2^n(sinx/2)(sinx/4)...(sinx/2^n)...}

=sinxsinx/2sinx/4....sinx/2^(n-1)..../

{2^n(sinx/2)(sinx/4)...(sinx/2^(n-1)(sinx/2^n)...

=(sinx)/(2^nsinx/2^n)=[{(sinx)/x}(x/2^n)]/(sinx/2^n).

As n→∞, x/2^n→0 (x/2^n)/(sinx/2^n) →1and the

whole limit is sinx/x.

You've followed me exactly up to your penultimate lime, Kevin. But there is a problem with your conclusion. This arises because as n→∞, 2^n→∞ and sin(x/2^n)→0. ∞x0 is indeterminate, but sinx/{(2^n)xsin(x/2^n)} has a definite limit, which if I am correct is sinx/x, or something else if I am wrong.

For clarity my 3rd line fom the bottom is [{(sinx)/x}(x/2^n)/sin(x/2^n)].

You've followed me exactly up to your penultimate lime, Kevin. But there is a problem with your conclusion. This arises because as n→∞, 2^n→∞ and sin(x/2^n)→0. ∞x0 is indeterminate, but sinx/{(2^n)xsin(x/2^n)} has a definite limit, which if I am correct is sinx/x, or something else if I am wrong.

For clarity my 3rd line fom the bottom is [{(sinx)/x}(x/2^n)/sin(x/2^n)].

None.

### 1 Answer

- KevinMLv 78 years agoFavorite Answer
First of all, you need to use parenthesis. I believe you mean cos(x/2) cos(x/4) ... In that case, your math seems ALMOST accurate. I get that this becomes:

lim sin x / (2^n sin (x/2^n) )

Which approaches 0 as n->infinity

Ed: You're right - as n-> infinity, 2^n sin (x/2^n) approaches

(-ln 2) x 2^(-n) cos (x/2^n) / -(ln 2) 2^(-n) = lim n-> infinity x cos (x/2^n) = x

So the limit becomes (sin x) / x. I even tried this out with a few values using Excel - neat problem!