# How to calculate this limit?#lim_(n->oo)prod_(k=1)^ncos(2^(k-1)x);x!=kpi#

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To calculate the limit ( \lim_{n \to \infty} \prod_{k=1}^n \cos\left(2^{k-1}x\right) ) given that ( x ) is not equal to ( k\pi ) for any integer ( k ), we can utilize the following steps:

- Recognize that the given limit involves an infinite product of cosine functions.
- Note that the cosine function oscillates between -1 and 1 as its argument varies.
- Observe that when ( x ) is not equal to ( k\pi ), the argument of each cosine function in the product will never be a multiple of ( \pi ), ensuring that the cosine function remains bounded between -1 and 1.
- Understand that the product of cosine functions will converge to zero if any of the cosine terms approach zero as ( n ) tends to infinity.
- Realize that since ( x ) is not equal to ( k\pi ) for any integer ( k ), the cosine terms will never be zero for any ( n ), and thus, the product of cosine functions will converge to a nonzero value as ( n ) tends to infinity.
- Therefore, the limit ( \lim_{n \to \infty} \prod_{k=1}^n \cos\left(2^{k-1}x\right) ) converges to a nonzero value, but its specific value depends on the particular value of ( x ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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