math question?

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  • 1 month ago

    with n→∞

    lim(n⁸+sin(12n+14))/(n¹²+14)=lim(n⁸)/(n¹²+14)  +lim(sin(12n+14)/(n¹²+14)

    So

    lim(n⁸)/(n¹²)=lim(1/n⁴)=0

    lim(sin(12n+14)/(n¹²))=lim(|1|/n¹²)=0

    0+0=0

  • Sean
    Lv 5
    1 month ago

    -1<=sin(u) <= 1

    so as n increases the value will differ negligibly from n^8/n^12 which soon is indistinguishably different from

    n^-4 which in the limit converges to  0

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