what is the limit n to infinite of

cos1*cos(1/2)*cos(1/4)*cos(1/8)*cos(1/16)*...*cos(1/2^n)

Since sinc(1) = cos(1/2)*cos(1/4)*cos(1/8)*cos(1/16)*...*cos(1/2^n)

the limit here is cos(1)sinc(1) = cos(1)sin(1) = 0.4546...

You can read about sinc(x) in various places. This function doesn't usually pop up in introductory calculus courses.

To find the limit as n approaches infinity of the expression cos(1) * cos(1/2) * cos(1/4) * cos(1/8) * ... * cos(1/2^n), we can use the concept of infinite products.

Step 1: Rewrite the expression

Let's rewrite the expression to make it easier to work with. We'll factor out the common term cosine(1):

cos(1) * cos(1/2) * cos(1/4) * cos(1/8) * ... * cos(1/2^n)
= cos(1) * (cos(1/2) * cos(1/4) * cos(1/8) * ... * cos(1/2^n))

Step 2: Use the formula for the infinite product of cosines

The formula for the infinite product of cosines is:

cos(x) * cos(x/2) * cos(x/4) * cos(x/8) * ... = sin(x) / x

Applying this formula to our expression:

= cos(1) * (sin(1) / 1)
= cos(1) * sin(1)

Step 3: Evaluate the limit

To find the limit as n approaches infinity, we need to evaluate the resulting expression:

lim(n→∞) cos(1) * sin(1)

cos(1) is a constant value, so we don't have to worry about its behavior. The limit only depends on sin(1).

Since sin(1) is also a constant value, the limit is simply the product of the two constant values:

lim(n→∞) cos(1) * sin(1) = cos(1) * sin(1)

Therefore, the limit of the given expression as n approaches infinity is cos(1) * sin(1).