If k is an integer such that Lim(n->infinity)[{cos((k Pi)/4)^n - cos((k Pi)/6)^n] = 0, then...?

A) k is divisible neither by 4 nor by 6;

B) k must be divisible by 12, but not necessarily by 24;

C) k must be divisible by 24;

D) either k is divisible by 24 or k is divisible neither by 4 nor by 6;

Kindly explain your answer...

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  • 1 decade ago
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    If k is not divisible by 4 or 6, then both cos (kπ/4) and cos (kπ/6) will have magnitude less than 1 and so their n'th powers will go to 0 as n goes to infinity. So that works.

    If k is divisible by either 4 or 6, but not both, then one term will have magnitude 1 and the other magnitude 0 in the limit, so that won't work.

    If k is divisible by both 4 and 6, it must be divisible by 12. If k is not divisible by 24, then kπ/4 will be an odd multiple of π and kπ/6 will be an even multiple of π, so we will get lim (n->∞) [(-1)^n - 1^n] which does not exist (the sequence is -2, 0, -2, 0, ....). If k is divisible by 24, both angles will be integer multiples of 2π, so every term will be 1^n - 1^n = 0. So this case also works.

    Hence either k is divisible by 24 or k is divisible by neither 4 nor 6. So (D) is the correct answer.

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