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Noah
Noah asked in Science & MathematicsMathematics · 5 years ago

Calculus limits question?

lim as n tends to infinity of (root(9n^2+2))/(root(4n^2+n+2))

3 Answers

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  • Brainard
    Lv 7
    5 years ago
    Favorite Answer

    (root(9n^2+2))/(root(4n^2+n+2))

    = sqrt[ (9n^2 + 2)/(4n^2 + n + 2)]

    = sqrt[ ( 9 + 2/n^2)/(4 + 1/n + 2/n^2)] dividing top and bottom by n^2 under sqrt

    As n→∞,, 2/n^2 , and 1/n → 0

    = sqrt(9/4)

    = 3/2

  • ?
    Lv 7
    5 years ago

    As n gets larger and larger, all terms become

    negligible except 9n^2 and 4n^2. So we essentially

    have √[ (9n^2)/(4n^2) ], which reduces to 3/2; that's

    the limit.

  • mizoo
    Lv 7
    5 years ago

    lim n→∞, (root(9n^2+2))/(root(4n^2+n+2)) = |3n|/|2n| = 3/2

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