$$ \left( 1 \right) 解：\lim_{n\rightarrow \infty} \left( \frac{1}{n^2}+\frac{2}{n^2}+\frac{3}{n^2}+\cdots +\frac{n-1}{n^2} \right) \begin{matrix}{l} =\lim_{n\rightarrow \infty} \frac{1}{n}\cdot \left( \frac{1}{n}+\frac{2}{n}+\frac{3}{n}+\cdots +\frac{n-1}{n} \right)\\ =\lim_{n\rightarrow \infty} \frac{1}{n}\sum_{k=1}^{n-1}{\frac{k}{n}}=\int_0^1{x}\mathrm{d}x=\frac{1}{2}x^2\mid_{0}^{1}\\ =\frac{1}{2}\cdot 1^2-0=\frac{1}{2}.\\ \end{matrix} \\ \left( 2 \right) 解：\lim_{n\rightarrow \infty} \frac{1^p+2^p+\cdots +n^p}{n^{p+1}}\begin{matrix}{l} =\lim_{n\rightarrow \infty} \frac{1}{n}\cdot \frac{1^p+2^p+\cdots +n^p}{n^p}\\ =\lim_{n\rightarrow \infty} \frac{1}{n}\cdot \left[ \left( \frac{1}{n} \right) ^p+\left( \frac{2}{n} \right) ^p+\cdots +\left( \frac{n}{n} \right) ^p \right]\\ =\lim_{n\rightarrow \infty} \frac{1}{n}\sum_{k=1}^n{\begin{array}{c} \begin{array}{c} \left( \frac{k}{n} \right) ^p=\int_0^1{x^p\mathrm{d}x}=\frac{1}{p+1}x^{p+1}\mid_{0}^{1}\\ \end{array}\\ \end{array}}\\ =\frac{1}{p+1}\cdot 1^{p+1}-0=\frac{1}{p+1}.\\ \end{matrix} $$