Find the limit of a Riemann Sum

Question

The function is $f(x) = 1-x^2$. I'm stuck as I can't factor the expression in the last line to find the limit.

Answer 1

The expression from your last line can be simplified: \begin{align} \frac{1}{n} \bigg[n-\frac{1}{n^2}\bigg(\frac{n(n+2)(2n+1)}{6}\bigg)\bigg] & = \bigg[\frac{n}{n}-\frac{n(n+2)(2n+1)}{6n^3}\bigg] \\[0.1in] & = 1-\frac{1}{6} \bigg( \frac{n+2}{n} \bigg)\bigg(\frac{2n+1}{n}\bigg) \\[0.1in] & = 1-\frac{1}{6} \bigg( 1+\frac{2}{n} \bigg)\bigg( 2+\frac{1}{n} \bigg) \end{align}

Answer 2

The last step should then be $$\lim_{n\to \infty }\{1-\frac{n(n+1)(2n+1)}{6n^3}\}=\lim_{n\to \infty }\{1-\frac{2n^3+\cdots}{6n^3}\}=1-\frac26$$