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Quasimodo · Answer 1 · 2023-11-21T15:34:45+0000

Given:

Required:

To find the limit value of the given function.

Step-by-step explanation:

$\begin{gathered} =\lim_(n\to\infty)[(1^2)/(n^3)+(2^2)/(n^3)+........+(n^2)/(n^3)] \\ =\lim_(n\to\infty)(1^2+2^2+3^2+.........+n^2)/(n^3) \\ =\operatorname{\lim}_(n\to\infty)(\sum_^n^2)/(n^3) \\ =\operatorname{\lim}_(n\to\infty)(1)/(n^3)*(n(n+1)(2n+1))/(6) \\ =\operatorname{\lim}_(n\to\infty)(n^3(1+(1)/(n))(2+(1)/(n)))/(6n^3) \end{gathered}$

Cancel out the same terms from the numerator and denominator.

$=\lim_(n\to\infty)((1+(1)/(n))(2+(1)/(n)))/(6)$

Now apply the limit.

$\begin{gathered} =\frac{(1+\frac{1}{\hat{\infty}})(2+(1)/(\infty))}{6} \\ =((1+0)(2+0))/(6) \\ =(2)/(6) \\ =(1)/(3) \end{gathered}$

Final Answer:

The limit value of the given function is

$\frac{1{}}{3}$
$\frac{1{}}{3}$

Evaluate the following limits. lim n -> ∞ sum i = 1 to n ((i ^ 2)/(n ^ 2)) * (1/n-example-1

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