2021 DHS JC2 MYE P1 Q2

Given that \(\mathrm{f}\) is a continuous function, explain, with the aid of a sketch, why the value of

\({\mathop{\lim}\limits_{n\to \infty }\,\frac{1}{n}\left[ \mathrm{f}\left( 0 \right)+\mathrm{f}\left( \frac{1}{n} \right)+\mathrm{f}\left( \frac{2}{n} \right)+\ldots +\mathrm{f}\left( 1-\frac{1}{n} \right) \right]}\)

is \(\displaystyle\displaystyle \displaystyle\int_{0}^{1}{\mathrm{f}\left( x \right)}\ \mathrm{d}x\).[3]

Hence, evaluate \({\mathop{\lim}\limits_{n\to \infty }\,\left[ \frac{1}{n}\left( 1+{{\mathrm{e}}^{\frac{2}{n}}}+{{\mathrm{e}}^{\frac{4}{n}}}+\ldots +{{\mathrm{e}}^{\left( 2-\frac{2}{n} \right)}} \right) \right]}\).[2]

For any given positive integer \(n\), explain whether \({\frac{1}{n}\left( 1+{{\mathrm{e}}^{\frac{2}{n}}}+{{\mathrm{e}}^{\frac{4}{n}}}+\ldots +{{\mathrm{e}}^{\left( 2-\frac{2}{n} \right)}} \right)}\) is larger or smaller than \({\mathop{\lim}\limits_{n\to \infty }\,\left[ \frac{1}{n}\left( 1+{{\mathrm{e}}^{\frac{2}{n}}}+{{\mathrm{e}}^{\frac{4}{n}}}+\ldots +{{\mathrm{e}}^{\left( 2-\frac{2}{n} \right)}} \right) \right]}\).[1]