Infi-1 4

Exercise

\begin{matrix} lim_{n \to \infty} \frac{n}{n^{2} + n - 9} = 0 \\ | \frac{n}{n^{2} + n - 9} - 0 | \underset{n > N \geq 3}{=} \frac{n}{n^{2} + n - 9} \underset{n > N \geq 9}{<} \frac{n}{n^{2}} = \frac{1}{n} < \frac{1}{N} \\ N = m a x (9, \frac{1}{ε}) \\ | \frac{n}{n^{2} + n - 9} - 0 | < \frac{1}{N} \underset{N \geq \frac{1}{ε}}{\leq} \frac{1}{\frac{1}{ε}} = ε \\ \forall ε > 0 \exists N = m a x (9, \frac{1}{ε}) : \forall n > N : | \frac{n}{n^{2} + n - 9} - 0 | < ε \end{matrix}

\begin{matrix} Prove: ∄ lim_{n \to \infty} (- 1)^{n} \\ If the finite limit exists: \\ \forall ε > 0 \exists N : \forall n > N : | (- 1)^{n} - L | < ε \\ Negation: \\ \exists ε > 0 : \forall N \exists n > N : | (- 1)^{n} - L | \geq ε \\ Does such ε exist? \\ Yes, for example: ε = \frac{1}{2} \\ If L \geq 0 : \\ \forall N \exists n = ⌈ 2 N ⌉ + 1 > N : 1 + L \geq \frac{1}{2} \\ If L < 0 : \\ \forall N \exists n = ⌈ 2 N ⌉ + 2 > N : 1 - L \geq \frac{1}{2} \\ s u p ((- 1)^{n}) = 1 ⟹ lim_{n \to \infty} (- 1)^{n} \neq \infty \\ i n f ((- 1)^{n}) = - 1 ⟹ lim_{n \to \infty} (- 1)^{n} \neq - \infty \\ Therefore, ∄ lim_{n \to \infty} (- 1)^{n} \end{matrix}

\begin{matrix} lim_{n \to \infty} a_{n} = L < 1 \\ Prove: \exists 0 < q < 1 : \exists N : \forall n > N : a_{n} < q \\ Let L < 0, ε = \frac{1}{2} > 0 \\ \exists N : \forall n > N : | a_{n} - L | < \frac{1}{2} \\ \exists N : \forall n > N : a_{n} < L + \frac{1}{2} \underset{L < 0}{<} \frac{1}{2} = q < 1 \\ ⟹ \exists 0 < q = \frac{1}{2} < 1 : \exists N : \forall n > N : a_{n} < q \\ Let L \geq 0, ε = \frac{1 - L}{2} \underset{L < 1}{>} 0 \\ \exists N : \forall n > N : | a_{n} - L | < \frac{1 - L}{2} \\ L - \frac{1 - L}{2} < a_{n} < L + \frac{1 - L}{2} \\ Let q = L + \frac{1 - L}{2} = \frac{L + 1}{2} \underset{\frac{1 - L}{2} < (1 - L)}{<} 1 \\ \exists N : \forall n > N : \frac{L - 1}{2} < a_{n} < q < 1 \\ L \geq 0 ⟹ q = \frac{L + 1}{2} \geq \frac{1}{2} > 0 \\ ⟹ \exists 0 < q = \frac{L + 1}{2} < 1 : \exists N : \forall n > N : a_{n} < q \end{matrix}

\begin{matrix} Prove or disprove: lim_{n \to \infty} a_{n} + b_{n} = L ⟹ a_{n}, b_{n} converges \\ Disproof: \\ a_{n} = (- 1)^{n} ⟹ ∄ lim_{n \to \infty} a_{n} \\ b_{n} = - (- 1)^{n} ⟹ ∄ lim_{n \to \infty} b_{n} \\ a_{n} + b_{n} = (- 1)^{n} - (- 1)^{n} = 0 ⟹ lim_{n \to \infty} a_{n} + b_{n} = 0 \\ Prove or disprove: lim_{n \to \infty} a_{n} + b_{n} = L, a_{n} diverges ⟹ b_{n} diverges \\ Proof: \\ Let lim_{n \to \infty} b_{n} = L_{b} \\ ⟹ lim_{n \to \infty} a_{n} = lim_{n \to \infty} a_{n} + b_{n} - b_{n} = lim_{n \to \infty} a_{n} + b_{n} - lim_{n \to \infty} b_{n} = L - L_{b} \\ ⟹ lim_{n \to \infty} a_{n} = L - L_{b} = L_{a} \\ But a_{n} diverges ⟹ Contradiction! \\ ⟹ b_{n} diverges \end{matrix}

\begin{matrix} lim_{n \to \infty} n (\sqrt{n^{2} + 1} - n) = lim_{n \to \infty} n (\sqrt{n^{2} + 1} - n) \cdot \frac{\sqrt{n^{2} + 1} + n}{\sqrt{n^{2} + 1} + n} = \\ = lim_{n \to \infty} n \frac{n^{2} + 1 - n^{2}}{\sqrt{n^{2} + 1} + n} = \frac{n}{\sqrt{n^{2} + 1} + n} = lim_{n \to \infty} \frac{n}{n (\sqrt{1 + \frac{1}{n^{2}}} + 1)} = \\ = lim_{n \to \infty} \frac{1}{\sqrt{1 + \frac{1}{n^{2}}} + 1} \\ lim_{n \to \infty} \frac{1}{n^{2}} = 0 ⟹ lim_{n \to \infty} \sqrt{1 + \frac{1}{n^{2}}} = \sqrt{1} ⟹ lim_{n \to \infty} \frac{1}{\sqrt{1} + 1} = \frac{1}{2} \end{matrix}

\begin{matrix} Prove or disprove: lim_{n \to \infty} a_{n} b_{n} = L ⟹ lim_{n \to \infty} a_{n}, b_{n} converge \\ Disproof: \\ a_{n} = b_{n} = (- 1)^{n} \\ lim_{n \to \infty} a_{n} b_{n} = (- 1)^{2 n} = 1 \\ But a_{n}, b_{n} diverge \\ Prove or disprove: lim_{n \to \infty} a_{n} b_{n} = L, a_{n} diverges ⟹ b_{n} diverges \\ Disproof: \\ a_{n} = (- 1)^{n}, b_{n} = 0 \\ lim_{n \to \infty} a_{n} b_{n} = (- 1)^{n} \cdot 0 = 0 \\ a_{n} diverges, but b_{n} converges \\ Prove or disprove: lim_{n \to \infty} a_{n} b_{n} = L, lim_{n \to \infty} b_{n} \neq 0, a_{n} diverges ⟹ b_{n} diverges \\ Proof: \\ Let lim_{n \to \infty} b_{n} = L_{b} \\ ⟹ lim_{n \to \infty} a_{n} = lim_{n \to \infty} \frac{a_{n} b_{n}}{b_{n}} = \frac{lim_{n \to \infty} a_{n} b_{n}}{lim_{n \to \infty} b_{n}} = \frac{L}{l_{b}} \\ ⟹ lim_{n \to \infty} a_{n} = \frac{L}{L_{b}} = L_{a} \\ But a_{n} diverges ⟹ Contradiction! \\ ⟹ b_{n} diverges \end{matrix}

\begin{matrix} lim_{n \to \infty} \frac{n^{3} - 6 n^{2} + 5 n + 1}{3 n^{3} + 2 n^{2} - 1} = lim_{n \to \infty} \frac{n^{3} (1 - \frac{6}{n} + \frac{5}{n^{2}} + \frac{1}{n^{3}})}{n^{3} (3 + \frac{2}{n} - \frac{1}{n^{3}})} = \\ = lim_{n \to \infty} \frac{1 - \overset{\to 0}{\frac{6}{n}} + \overset{\to 0}{\frac{5}{n^{2}}} + \overset{\to 0}{\frac{1}{n^{3}}}}{3 + \underset{\to 0}{\frac{2}{n}} - \underset{\to 0}{\frac{1}{n^{3}}}} = \frac{1}{3} \end{matrix}