Infi-1 3

\begin{matrix} Number L is a limit of sequence a_{n} (a_{n} approaches L) if and only if: \\ \forall ε > 0 \exists n_{ε} : \forall n > n_{ε} : | a_{n} - L | < ε \\ The limit notation is as following: \\ a_{n} \to L or lim_{n \to \infty} a_{n} = L \end{matrix}

Prove: lim_{n \to \infty} \frac{1}{4 n} = 0

\begin{matrix} | \frac{1}{4 n} - 0 | = \frac{1}{4 n} < \frac{1}{4 n_{ε}} \\ n_{ε} = \frac{1}{4 ε} \\ | \frac{1}{4 n} - 0 | < \frac{1}{4 n_{ε}} = \frac{1}{4 \frac{1}{4 ε}} = ε \\ ⟹ \forall ε > 0 \exists n_{ε} = \frac{1}{4 ε} : \forall n > n_{ε} : | \frac{1}{4 n} - 0 | < ε \\ ⟹ lim_{n \to \infty} \frac{1}{4 n} = 0 \end{matrix}

Prove: lim_{n \to \infty} \frac{n^{2} + 1}{n^{2}} = 1

\begin{matrix} | \frac{n^{2} + 1}{n^{2}} - 1 | = | \frac{n^{2} + 1 - n^{2}}{n^{2}} | = \frac{1}{n^{2}} < \frac{1}{n_{ε}^{2}} \\ n_{ε} = \sqrt{\frac{1}{ε}} \\ | \frac{n^{2} + 1}{n^{2}} - 1 | < \frac{1}{n_{ε}^{2}} = \frac{1}{{\sqrt{\frac{1}{ε}}}^{2}} = ε \\ ⟹ \forall ε > 0 \exists n_{ε} = \frac{1}{\sqrt{ε}} : \forall n > n_{ε} : | \frac{n^{2} + 1}{n^{2}} - 1 | < ε \\ ⟹ lim_{n \to \infty} \frac{n^{2} + 1}{n^{2}} = 1 \end{matrix}

Prove: lim_{n \to \infty} \frac{2 n - 1}{n} = 2

\begin{matrix} | \frac{2 n - 1}{n} - 2 | = | \frac{2 n - 1 - 2 n}{n} | = | \frac{- 1}{n} | = \frac{1}{n} < \frac{1}{n_{ε}} \\ n_{ε} = ⌈ \frac{1}{ε} ⌉ + 1 \\ | \frac{2 n - 1}{n} - 2 | < \frac{1}{n_{ε}} = \frac{1}{⌈ \frac{1}{ε} ⌉ + 1} < \frac{1}{⌈ \frac{1}{ε} ⌉} \leq \frac{1}{\frac{1}{ε}} = ε \\ ⟹ \forall ε > 0 \exists n_{ε} = ⌈ \frac{1}{ε} ⌉ + 1 : \forall n > n_{ε} : | \frac{2 n - 1}{n} - 2 | < ε \\ ⟹ lim_{n \to \infty} \frac{2 n - 1}{n} = 2 \end{matrix}

Prove: lim_{n \to \infty} \frac{1}{n + 7} = 0

\begin{matrix} | \frac{1}{n + 7} - 0 | = \frac{1}{n + 7} < \frac{1}{n_{ε} + 7} \\ n_{ε} = ⌈ \frac{1}{ε} - 7 ⌉ + 10 \\ | \frac{1}{n + 7} - 0 | < \frac{1}{n_{ε} + 7} = \frac{1}{⌈ \frac{1}{ε} - 7 ⌉ + 17} \leq \frac{1}{\frac{1}{ε} + 10} < \frac{1}{\frac{1}{ε}} = ε \\ ⟹ \forall ε > 0 \exists n_{ε} = ⌈ \frac{1}{ε} - 7 ⌉ + 10 : \forall n > n_{ε} : | \frac{1}{n + 7} - 0 | < ε \\ ⟹ lim_{n \to \infty} \frac{1}{n + 7} = 0 \end{matrix}

Prove: lim_{n \to \infty} \frac{n^{2} - 1}{3 n^{2} + n + 1} = \frac{1}{3}

\begin{matrix} | \frac{n^{2} - 1}{3 n^{2} + n + 1} - \frac{1}{3} | = | \frac{3 n^{2} - 3 - (3 n^{2} + n + 1)}{9 n^{2} + 3 n + 3} | = | \frac{- n - 4}{9 n^{2} + 3 n + 3} | = \\ = \frac{n + 4}{9 n^{2} + 3 n + 3} \leq \frac{n + 4 n}{9 n^{2} + 3 n + 3} < \frac{5 n}{9 n^{2}} = \frac{5}{9 n} < \frac{5}{5 n} < \frac{1}{n_{ε}} \\ n_{ε} = ⌈ \frac{1}{ε} ⌉ \\ | \frac{n^{2} - 1}{3 n^{2} + n + 1} - \frac{1}{3} | < \frac{1}{n_{ε}} = \frac{1}{⌈ \frac{1}{ε} ⌉} \leq \frac{1}{\frac{1}{ε}} = ε \\ ⟹ \forall ε > 0 \exists n_{ε} = ⌈ \frac{1}{ε} ⌉ : \forall n > n_{ε} : | \frac{n^{2} - 1}{3 n^{2} + n + 1} - \frac{1}{3} | < ε \\ ⟹ lim_{n \to \infty} \frac{n^{2} - 1}{3 n^{2} + n + 1} = \frac{1}{3} \end{matrix}

Prove: lim_{n \to \infty} a_{n} = 1 ⟹ \exists n_{ε} : \forall n > n_{ε} : \exists a_{n} > \frac{1}{2}

\begin{matrix} lim_{n \to \infty} a_{n} = 1 ⟹ \forall ε > 0 : \exists n_{ε} : \forall n > n_{ε} : | a_{n} - 1 | < ε \\ Let ε = \frac{1}{2} \\ \exists n_{ε} : \forall n > n_{ε} : | a_{n} - 1 | < \frac{1}{2} \\ ⟹ \exists n_{ε} : \forall n > n_{ε} : \frac{1}{2} < a_{n} < \frac{3}{2} \end{matrix}

Prove: lim_{n \to \infty} a_{n} = L > 0 ⟹ \exists n_{ε} : \forall n > n_{ε} : \exists a_{n} < \frac{11}{10} L

\begin{matrix} lim_{n \to \infty} a_{n} = L ⟹ \forall ε > 0 : \exists n_{ε} : \forall n > n_{ε} : | a_{n} - L | < ε \\ Let ε = \frac{L}{10} \\ \exists n_{ε} : \forall n > n_{ε} : | a_{n} - L | < \frac{L}{10} \\ ⟹ \exists n_{ε} : \forall n > n_{ε} : \frac{9}{10} L < a_{n} < \frac{11}{10} L \end{matrix}