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Math Help - Find d such that function tends to C as n tends to infinity

  1. #1
    Junior Member
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    Find d such that function tends to C as n tends to infinity

    Hi, I need help on this question. I have an answer to it but i'm not sure its right. Here is goes,

    Find d such that

    {4^n}/ ({\frac{1}{3^n}})^{-d}\rightarrow C as n \rightarrow \infty

    Where C is a non-zero constant.

    I rearranged the problem to get (\frac{4}{3^d})^n and then if
    d ≤ 1 then {4^n}/ ({\frac{1}{3^n}})^{-d} diverges as n \rightarrow \infty
    d \geq2 then {4^n}/ ({\frac{1}{3^n}})^{-d}\rightarrow 0 as as n \rightarrow \infty

    So only value of d that converges to a non zero constant is the value of d that makes {4^n}/ ({\frac{1}{3^n}})^{-d} = 1. Is that right? so d = \frac{ln4}{ln3}?

    Please help, coz i'm not sure what i'm doing.
    Thanks
    Katy
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  2. #2
    Senior Member
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    Quote Originally Posted by harkapobi View Post
    I rearranged the problem to get (\frac{4}{3^d})^n and then if
    d ≤ 1 then {4^n}/ ({\frac{1}{3^n}})^{-d} diverges as n \rightarrow \infty
    d \geq2 then {4^n}/ ({\frac{1}{3^n}})^{-d}\rightarrow 0 as as n \rightarrow \infty
    what happened when 1 < d < 2?


    So only value of d that converges to a non zero constant is the value of d that makes {4^n}/ ({\frac{1}{3^n}})^{-d} = 1. Is that right? so d = \frac{ln4}{ln3}?
    yes
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