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lim(n->oo)(1^2+2^2+3^2++n^2)/(n^3) is eq...

`lim_(n->oo)(1^2+2^2+3^2++n^2)/(n^3)` is equal to a. 1 b . c. 1/3 d. 0

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To solve the limit \( \lim_{n \to \infty} \frac{1^2 + 2^2 + 3^2 + \ldots + n^2}{n^3} \), we can follow these steps: ### Step 1: Use the formula for the sum of squares The sum of the squares of the first \( n \) natural numbers is given by the formula: \[ 1^2 + 2^2 + 3^2 + \ldots + n^2 = \frac{n(n + 1)(2n + 1)}{6} \] ...
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Knowledge Check

  • lim_(n to oo) (1^(2)+2^(2)+3^(2)+…+n^(2))/(n^(3)) is equal to

    A
    `(1)/(2)`
    B
    `(2)/(3)`
    C
    `(1)/(3)`
    D
    `(1)/(6)`

  • lim_(n to oo) ((2n^(2)-3)/(2n^(2)-n+1))^((n^(2)-1)/(n)) is equal to

    A
    `(1)/(sqrt(e ))`
    B
    `sqrt(e )`
    C
    e
    D
    `(1)/(e )`

  • lim_(n rarr oo) II_(n=2)^(n) (1 - (3)/(n (n+2))) is equal to

    A
    1
    B
    4
    C
    `(1)/(4)`
    D
    `(3)/(4)`

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