10.25 Supremum and the Completeness Axiom

Week 2 part 2.pdf

1. Prove \(\begin{aligned}\lim_{n\rightarrow\infty}\frac{2n^2}{n^3+3}=0\end{aligned}\)​

   For \(\varepsilon>0,\exists n_0\in\mathbb{N},n>\frac{2}{\varepsilon}\) such that for \(n\geq n_0\) such that \(|\frac{2n^2}{n^3+3}-0|=\frac{2n^2}{n^3+3}<\frac{2n^2}{n^3}=\frac{2}{n}<\frac{2}{n_0}<\varepsilon\)​

2. Prove \(\begin{aligned}\lim_{n\rightarrow\infty}\frac{2n+1}{5n+4}=\frac25\end{aligned}\)​

   For \(\varepsilon>0,\exists n_0\in\mathbb{N},n>\frac{3}{25\cdot\varepsilon}\) such that for \(n\geq n_0\) such that \(|\frac{2n+1}{5n+4}-\frac25|=\frac{3}{25n}\leq\frac{3}{25n_0}<\varepsilon\)​

3. Prove \(\begin{aligned}\lim_{n\rightarrow\infty}\sqrt{n^2+1}-n=0\end{aligned}\)

   For \(\varepsilon>0,\exists n_0\in\mathbb{N},n>\frac{1}{2\varepsilon}\) such that for \(n\geq n_0\) such that \(|\sqrt{n^2+1}-n-0|=\sqrt{n^2+1}-n=(\sqrt{n^2+1}-n)\cdot\frac{\sqrt{n^2+1}+n}{\sqrt{n^2+1}+n}=\frac{n^2+1-n^2}{\sqrt{n^2+1}+n}=\frac{1}{\sqrt{n^2+1}+n}\leq\frac{1}{n+n}=\frac{1}{2n}<\varepsilon\)​

4. Exercise: Show that for all \(p \geq 1\), \(\begin{aligned}\lim_{n \to \infty} \frac{1}{n^p} = 0\end{aligned}\)

   Now, \(\forall n>n_0\) we have \(n^{p}>n>n_0>\frac{1}{\varepsilon}\Rightarrow n^{p}>\frac{1}{\varepsilon}\Rightarrow\varepsilon>\frac{1}{n^{p}}\Rightarrow\frac{1}{n^{p}}<\varepsilon\Rightarrow|\frac{1}{n^{p}}|<\varepsilon\)​

5. \(\begin{aligned}\lim_{n \to \infty} \frac{\sin(n^2)}{3 \sqrt{n}} = 0\end{aligned}\)

   For \(\varepsilon>0,\exists n_0\in\mathbb{N},n_0>\frac{1}{9\varepsilon^2}\) such that for \(n\geq n_0\) such that \(|\frac{\sin(n^2)}{3\sqrt{n}}-0|=\frac{\sin(n^2)}{3\sqrt{n}}\leq\frac{1}{3\sqrt{n}}<\frac{1}{3\sqrt{n_0}}<\varepsilon\)​

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