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Math Help - Compute the limit (3^n)/(2^n)+1

  1. #1
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    Compute the limit (3^n)/(2^n)+1

    Compute the limit of the sequence

    (3^n)/ ((2^n)+1)

    I have simplified a much harder problem to this limit and i am having a mind blank... helpplease? cheers
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  2. #2
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    The nth term tends to (3/2)^n.
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  3. #3
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    Quote Originally Posted by sebjory View Post
    Compute the limit of the sequence

    (3^n)/ ((2^n)+1)

    I have simplified a much harder problem to this limit and i am having a mind blank... helpplease? cheers
    \frac{3^n}{2^n + 1} \geq \frac{3^n}{2^n + 2^n} = \tfrac{1}{2}\left( \tfrac{3}{2} \right)^n \to \infty
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  4. #4
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    Quote Originally Posted by sebjory View Post
    Compute the limit of the sequence

    (3^n)/ ((2^n)+1)

    I have simplified a much harder problem to this limit and i am having a mind blank... helpplease? cheers
    Could also use L'Hoptial's rule on

    \lim_{x \to \infty} \frac{3^x}{2^x+1} = \lim_{x \to \infty} \frac{3^x \ln 3}{2^x \ln 2} = \frac{\ln 3}{\ln 2} \lim_{x \to \infty} \left( \frac{3}{2} \right)^x \to \infty as seen in the ThePerfectHacker post.
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  5. #5
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    Hello, sebjory!

    Yet another approach . . .


    \lim_{n\to\infty} \frac{3^n}{2^n + 1}

    Divide top and bottom by 3^n\!:\;\;\frac{\frac{1}{3^n}\cdot3^n}{\frac{1}{3^  n}(2^n+1)} \;=\;\frac{1}{\frac{2^n}{3^n} + \frac{1}{3^n}}


    Therefore: . \lim_{n\to\infty}\left[\frac{1}{\left(\frac{2}{3}\right)^n + \frac{1}{3^n}}\right] \;=\; \frac{1}{0+0} \;=\;\infty

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  6. #6
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    thanks a hell of a lot guys that was exactly what i was looking for. Think i might use Sorobans solution nice and elegant. cant believe i didnt see it
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, sebjory!

    Yet another approach . . .

    \frac{1}{0+0} \;=\;\infty
    Soroban! Nooo!
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