Exam 2023B (A)

1

\begin{matrix} Let x > 3 \\ Prove: \frac{2 \ln (x - 2)}{\sqrt{x + 1} - 2} > \frac{4 \sqrt{x + 1}}{x - 2} \\ Proof: \\ Let f (x) = 2 \ln (x - 2) \\ f (3) = 0 \\ f^{'} (x) = \frac{2}{x - 2} \\ Let g (x) = \sqrt{x + 1} - 2 \\ g^{'} (x) = \frac{1}{2 \sqrt{x + 1}} \\ g (3) = 0 \\ By the Lagrange theorem: \exists c \in (3, x) : \frac{f (x)}{g (x)} = \frac{f (x) - f (3)}{g (x) - g (3)} = \frac{f^{'} (c)}{g^{'} (c)} \\ Let h (x) = \frac{f^{'} (x)}{g^{'} (x)} = \frac{4 \sqrt{x + 1}}{x - 2} \\ h^{'} (x) = \frac{\frac{2 x - 4}{\sqrt{x + 1}} - 4 \sqrt{x + 1}}{(x - 2)^{2}} = \frac{2 x - 4 - 4 x - 4}{\sqrt{x + 1} (x - 2)^{2}} = \frac{- 2 x - 8}{\sqrt{x + 1} (x - 2)^{2}} \\ \forall x > 3 : h^{'} (x) < 0 ⟹ \forall 3 < c < x : h (c) > h (x) \\ ⟹ \frac{f (x)}{g (x)} > h (x) ⟹ \frac{2 \ln (x - 2)}{\sqrt{x + 1} - 2} > \frac{4 \sqrt{x + 1}}{x - 2} \end{matrix}

\begin{matrix} Let f be a function on I \\ f is called continuous in equal measure on I if \\ \forall {a_{n}}, {b_{n}} \subseteq I : a_{n} - b_{n} \to 0 ⟹ f (a_{n}) - f (b_{n}) \to 0 \end{matrix}

\begin{matrix} Let f be a function \\ Let f be continuous on [a, b] \\ Prove: f is continuous in equal measure on [a, b] \\ Proof: \\ Let {a_{n}}, {b_{n}} \subseteq [a, b] : a_{n} - b_{n} \to 0 \\ Let f (a_{n}) - f (b_{n}) ↛ 0 \\ ⟹ \exists ε > 0 : \exists n_{k} : \forall k : | f (a_{n_{k}}) - f (b_{n_{k}}) | > ε \\ {a_{n_{k}} - b_{n_{k}}} \subseteq {a_{n} - b_{n}} ⟹ a_{n_{k}} - b_{n_{k}} \to 0 \\ {a_{n_{k}}} \subseteq {a_{n}} \subseteq [a, b] ⟹ \exists {a_{n_{k_{j}}}} \to L \in [a, b] \\ {b_{n_{k_{j}}}} \subseteq {b_{n}} \subseteq [a, b] \\ a_{n_{k_{j}}} - b_{n_{k_{j}}} \to 0 ⟹ b_{n_{k_{j}}} \to L \\ ⟹ f (a_{n_{k_{j}}}) \to f (L), f (b_{n_{k_{j}}}) \to f (L) \\ ⟹ f (a_{n_{k_{j}}}) - f (b_{n_{k_{j}}}) \to 0 \\ {f (a_{n_{k_{j}}}) - f (b_{n_{k_{j}}})} \subseteq {f (a_{n_{k}}) - f (b_{n_{k}})} \\ ⟹ \forall j \in N : | f (a_{n_{k_{j}}}) - f (b_{n_{k_{j}}}) | \geq ε - Contradiction! \\ ⟹ f (a_{n}) - f (b_{n}) \to 0 \end{matrix}

\begin{matrix} Show: f (x) = x^{2} is not continuous in equal measure on [1, \infty) \\ Solution: \\ Let a_{n} = n + \frac{1}{n} \\ Let b_{n} = n \\ a_{n} - b_{n} = \frac{1}{n} \to 0 \\ f (a_{n}) - f (b_{n}) = {(n + \frac{1}{n})}^{2} - n^{2} = n^{2} + 2 + \frac{1}{n^{2}} - n^{2} = 2 + \frac{1}{n^{2}} \to 2 \\ ⟹ f (a_{n}) - f (b_{n}) ↛ 0 \end{matrix}

\begin{matrix} Let a_{n} = \sum_{k = n}^{3 n} \frac{1}{k} \end{matrix}

\begin{matrix} Prove: a_{n} converges \\ Proof: \\ a_{n + 1} - a_{n} = \sum_{k = n + 1}^{3 n + 3} \frac{1}{k} - \sum_{k = n}^{3 n} \frac{1}{k} = \frac{1}{3 n + 1} + \frac{1}{3 n + 2} + \frac{1}{3 n + 3} - \frac{1}{n} < \frac{3}{3 n} - \frac{1}{n} = 0 \\ ⟹ a_{n + 1} < a_{n} ⟹ a_{n} is monotonically descending \\ a_{n} \geq \sum_{k = n}^{3 n} \frac{1}{3 n} = \frac{2 n}{3 n} = \frac{2}{3} \\ ⟹ a_{n} is lower-bounded ⟹ a_{n} converges \end{matrix}

\begin{matrix} Prove: a_{n} \to L \geq \frac{2}{3} \\ Proof: \\ a_{n} \geq \frac{2}{3} ⟹ L \geq \frac{2}{3} \end{matrix}

\begin{matrix} Copy of 2024(A) \end{matrix}

\begin{matrix} Prove: (a r c c o t x)^{'} = - \frac{1}{1 + x^{2}} \\ Proof: \\ (f^{- 1})^{'} = \frac{1}{f^{'} (f^{- 1} (x))} \\ Let f (x) = \cot x \\ f^{'} (x) = {(\frac{\cos x}{\sin x})}^{'} = \frac{- \sin^{2} x - \cos^{2} x}{\sin^{2} x} = - 1 - \cot^{2} x \\ ⟹ f^{'} (f^{- 1} (x)) = - 1 - \cot^{2} (a r c c o t x) = - 1 - x^{2} \\ (f^{- 1} (x))^{'} = \frac{1}{- 1 - x^{2}} = - \frac{1}{1 + x^{2}} \end{matrix}

\begin{matrix} Find: lim_{x \to 0^{+}} \frac{1}{x} (\frac{1}{x^{x}} - 1) \\ Solution: \\ \frac{1}{x} (\frac{1}{x^{x}} - 1) = \frac{x^{- x} - 1}{x} = \frac{e^{- x \ln x} - 1}{x} \\ lim_{x \to 0^{+}} x \ln x = \dots = 0 ⟹ e^{- x \ln x} \to 1 \\ ⟹ lim_{x \to 0^{+}} \frac{e^{- x \ln x} - 1}{x} \overset{L}{=} lim_{x \to 0^{+}} - (x \ln x)^{'} e^{- x \ln x} = - lim_{x \to 0^{+}} (\ln x + 1) \cdot lim_{x \to 0^{+}} e^{- x \ln x} = \\ = - (- \infty) \cdot 1 = \infty \\ ⟹ lim_{x \to 0^{+}} \frac{1}{x} (\frac{1}{x^{x}} - 1) = \infty \end{matrix}