자연계열 1번

[1-1 풀이]$\displaystyle \lim_{x\to n+0}f(x)=pn^2+kqn$$\displaystyle \lim_{x\to n-0}f(x)=pn^2-1+k(qn-1)$

함수 $\displaystyle f(x)$가 $\displaystyle x=n$에서 연속이 되려면 $\displaystyle \lim_{x\to n-0}f(x)=f(n)=\lim_{x\to n+0}f(x)$를 만족해야 한다.따라서 $\displaystyle pn^2+kqn=pn^2-1+k(qn-1)$이고 $\displaystyle 0=-1-k$그러므로 $\displaystyle k=-1$.

[1-2 풀이1]$\displaystyle \lim_{h\to+0}\frac{f(n+h)-f(n)}h=\lim_{h\to+0}\frac{(pn^2-qn)-(pn^2-qn)}h=0$$\displaystyle \lim_{h\to-0}\frac{f(n+h)-f(n)}h=\lim_{h\to-0}\frac{((pn^2-1)-(qn-1))-(pn^2-qn)}h=0$

[1-3 풀이]$\displaystyle \begin{aligned}\int_0^1[nx^2]dx&=\sum_{k=0}^{n-1}\int_{\sqrt{\frac kn}}^{\sqrt{\frac{k+1}n}}k\,dx=\sum_{k=0}^{n-1}k\left(\sqrt{\frac{k+1}n}-\sqrt{\frac kn}\right)\\&=\frac1{\sqrt n}\{(\sqrt2-1)+2(\sqrt3-\sqrt2)+3(\sqrt4-\sqrt3)+\cdots+(n-1)(\sqrt n-\sqrt{n-1})\}\\&=\frac1{\sqrt n}\{(n-1)\sqrt n-\sqrt{n-1}-\sqrt{n-2}-\cdots-\sqrt2-1\}\\&=n-\frac1{\sqrt n}\sum_{k=1}^n\sqrt k\end{aligned}$따라서 $\displaystyle \frac1n\int_0^1[nx^2]dx=1-\frac1n\sum_{k=1}^n\sqrt{\frac kn}$이고$\displaystyle \begin{aligned}\lim_{n\to\infty}\frac1n\int_0^1[nx^2]dx&=1-\lim_{n\to\infty}\frac1n\sum_{k=1}^n\sqrt{\frac kn}\\&=1-\int_0^1\sqrt t\,dt=1-\frac23=\frac13\end{aligned}$

[1-4 풀이]제시문 <나>에 의하여 구간 $\displaystyle [0,1]$에서 함수 $\displaystyle f(x)$의 역함수 $\displaystyle g(x)$가 존재하고$\displaystyle f(0)=g(0)=0,\ f(1)=g(1)=1$따라서$\displaystyle \begin{aligned}\int_0^1[nf(x)]dx&=\sum_{k=0}^{n-1}\int_{g\left(\frac kn\right)}^{g\left(\frac{k+1}n\right)}k\,dx=\sum_{k=0}^{n-1}k\left\{g\left(\frac{k+1}n\right)-g\left(\frac kn\right)\right\}\\&=ng\left(\frac nn\right)-g\left(\frac nn\right)-g\left(\frac{n-1}n\right)-g\left(\frac{n-2}n\right)-\cdots-g\left(\frac2n\right)-g\left(\frac1n\right)\\&=n-\sum_{k=1}^ng\left(\frac kn\right)\end{aligned}$

따라서 $\displaystyle \int_0^1\frac{[nf(x)]}{n}dx=1-\frac1n\sum_{k=1}^ng\left(\frac kn\right)$이고$\displaystyle \begin{aligned}\lim_{n\to\infty}\int_0^1\frac{[nf(x)]}{n}dx&=1-\lim_{n\to\infty}\frac1n\sum_{k=1}^ng\left(\frac kn\right)\\&=1-\int_0^1g(t)dt=\int_0^1f(x)dx\end{aligned}$그러므로 $\displaystyle \lim_{n\to\infty}\int_0^1\left(\frac{[nf(x)]}{n}-c\right)dx=0$