Question

Evaluate the following limits. lim n -> ∞ sum i = 1 to n ((i ^ 2)/(n ^ 2)) * (1/n) (Picture of equation for clarification)

Answer 1

Given:

Required:

To find the limit value of the given function.

Explanation:

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

Cancel out the same terms from the numerator and denominator.

$=\lim_{n\to\infty}\frac{(1+\frac{1}{n})(2+\frac{1}{n})}{6}$

Now apply the limit.

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

Final Answer:

The limit value of the given function is

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