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Math Help - converges

  1. #1
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    converges

    ihi........ i have ths question where i have to decide whether it converges and if it does i need to state its limit.

    an= 3n^4 -12 / 5n^3 + 8

    i would devide it all by n giving 3n^3 - 12/n / 5n^2 + 8/n

    but i do not know if this is the right approce andwhere to go after this.

    please help me in getting in to the right direction

    thanx
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  2. #2
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    Quote Originally Posted by redieeee_babieeee View Post
    ihi........ i have ths question where i have to decide whether it converges and if it does i need to state its limit.

    an= 3n^4 -12 / 5n^3 + 8

    i would devide it all by n giving 3n^3 - 12/n / 5n^2 + 8/n

    but i do not know if this is the right approce andwhere to go after this.

    please help me in getting in to the right direction

    thanx
    Is
    a_n = \frac{3n^4 -12 }{5n^3 + 8} ?

    If so, then no.
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  3. #3
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    yes lol it is that
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  4. #4
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    Hello, redieeee_babieeee!

    Decide whether it converges; if it does, state its limit.

    . . a_n\:=\: \frac{3n^4 -12}{5n^3 + 8}

    Divide top and bottom by the highest power of n in the denominator, . n^3

    \lim_{n\to\infty}\frac{\dfrac{3n^4}{n^3} - \dfrac{12}{n^3}}{\dfrac{5n^3}{n^3} + \dfrac{8}{n^3}} \;\;=\;\;\lim_{n\to\infty}\frac{3n - \dfrac{12}{n^3}}{5 + \dfrac{8}{n^3}} \;\;=\;\;\frac{\infty - 0}{5 + 0} \:\;=\;\:\infty . . . diverges

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