Infi-1 11

Exercise

\begin{matrix} Let f (x) > 0 \\ \forall x \in R : \exists f (x) \\ lim_{x \to 0} f (x) = 0 \\ Prove: \\ 1. lim_{n \to \infty} f (\frac{1}{n}) = 0 \\ 2. lim_{n \to \infty} f (f (\frac{1}{n})) = 0 \\ Proof: \\ lim_{x \to 0} f (x) = 0 ⟹ \forall x_{n} : [x_{n} \to 0, x_{n} \neq 0 ⟹ f (x_{n}) \to 0] \\ Let x_{n} = \frac{1}{n}, x_{n} \to 0, x_{n} \neq 0 \\ ⟹ f (x_{n}) = f (\frac{1}{n}) \to 0 \\ Let x_{n} = f (\frac{1}{n}), x_{n} \to 0, x_{n} \neq 0 \\ ⟹ f (x_{n}) = f (f (\frac{1}{n})) \to 0 \end{matrix}

\begin{matrix} lim_{x \to x_{0}^{+}} f (x) = L ⟺ \forall x_{n} : [x_{n} \to x_{0}, x_{n} > x_{0} ⟹ f (x_{n}) \to L] \\ lim_{x \to x_{0}^{-}} f (x) = L ⟺ \forall x_{n} : [x_{n} \to x_{0}, x_{n} < x_{0} ⟹ f (x_{n}) \to L] \end{matrix}

\begin{matrix} lim_{x \to 1^{-}} \frac{1 - x}{| x - 1 |} = 1 \\ Proof: \\ Let x_{n} \to 1, x_{n} < 1 \\ ⟹ lim_{n \to \infty} f (x_{n}) = lim_{n \to \infty} \frac{1 - x_{n}}{| x_{n} - 1 |} \underset{x_{n} < 1}{=} lim_{n \to \infty} \frac{1 - x_{n}}{1 - x_{n}} = lim_{n \to \infty} 1 = 1 \end{matrix}

\begin{matrix} lim_{x \to 0} \frac{x \sin x}{\sqrt{2} - \sqrt{1 + \cos x}} = lim_{x \to 0} \frac{x \sin x (\sqrt{2} + \sqrt{1 + \cos x})}{1 - \cos x} = \\ = lim_{x \to 0} x^{2} \frac{\frac{\sin x}{x} (\sqrt{2} + \sqrt{1 + \cos x})}{x^{2} \frac{1 - \cos x}{x^{2}}} = lim_{x \to 0} \underset{\to 1}{\underset{⏟}{\frac{\sin x}{x}}} \cdot \underset{\to 2}{\underset{⏟}{\frac{1}{\frac{1 - \cos x}{x^{2}}}}} \cdot (\sqrt{2} + \underset{\to \sqrt{2}}{\underset{⏟}{\sqrt{1 + \cos x}}}) = \\ = 1 \cdot 2 \cdot 2 \sqrt{2} = 4 \sqrt{2} \end{matrix}

\begin{matrix} lim_{x \to x_{0}} f (x) = f (x_{0}) \end{matrix}

\begin{matrix} Prove or disprove: \\ f, g not continuous at x_{0} \\ 1. f + g not continuous at x_{0} \\ 2. f \cdot g not continuous at x_{0} \\ Disproof for 1. \\ Let f (x) = D (x) = {\begin{cases} 1 & x \in Q \\ 0 & otherwise \end{cases} \\ Let g (x) = - D (x) = {\begin{cases} - 1 & x \in Q \\ 0 & otherwise \end{cases} \\ (f + g) (x) = 0 which is continuous \\ Disproof for 2. \\ Let f (x) = D (x) \\ Let g (x) = D^{'} (x) = {\begin{cases} 0 & x \in Q \\ 1 & otherwise \end{cases} \\ (f \cdot g) (x) = 0 which is continuous \end{matrix}

\begin{matrix} Let f \\ \forall x \in R : \exists f (x) \\ \forall a, b \in R : f (a + b) = f (a) + f (b) \\ f is continuous at 0 \\ Prove: \\ 1. f (0) = 0 \\ 2. f is continuous on R \\ Proof: \\ Let a = b = 0 \\ f (a + b) = f (a) + f (b) = f (0) = f (0) + f (0) \\ ⟹ f (0) = 0 \\ \forall a \in R : lim_{x \to a} f (x) = f (a) \\ Let x_{n} \to a, x_{n} \neq a \\ Let b = - a \\ lim_{n \to \infty} f (x_{n}) = lim_{n \to \infty} f (x_{n} - a + a) = \\ = lim_{n \to \infty} (f (\underset{\to 0}{\underset{⏟}{x_{n} - a}}) + f (a)) = f (0) + f (a) = 0 + f (a) = f (a) \\ ⟹ f is continuous at a ⟹ f is continuous on R \end{matrix}

\begin{matrix} lim_{x \to 0} \frac{\ln (1 + x)}{x} = 1 \\ lim_{x \to 0} \frac{a^{x} - 1}{x} = \ln a \end{matrix}

\begin{matrix} lim_{x \to 0} \frac{\ln (1 + x) - \ln (1 - x)}{x} = lim_{x \to 0} (\frac{\ln (1 + x)}{x} - \frac{\ln (1 - x)}{x}) = \\ = lim_{x \to 0} \frac{\ln (1 + x)}{x} + \frac{\ln (1 - x)}{- x} = 1 + 1 = 2 \end{matrix}

\begin{matrix} lim_{x \to 0} \frac{e^{2 x} - 1}{e^{3 x} - 1} = lim_{x \to 0} \frac{\overset{\to 1}{\overset{⏞}{\frac{e^{2 x} - 1}{2 x}}} \cdot 2 x}{\underset{\to 1}{\underset{⏟}{\frac{e^{3 x} - 1}{3 x}}} \cdot 3 x} = \frac{2}{3} \end{matrix}

\begin{matrix} \exists lim_{x \to a} f (x) \in R \\ lim_{x \to a} f (x) \neq f (a) \end{matrix}

\begin{matrix} \exists lim_{x \to a^{-}} f (x) \in R, \exists lim_{x \to a^{+}} f (x) \in R \\ lim_{x \to a^{-}} f (x) \neq lim_{x \to a^{+}} f (x) \end{matrix}

\begin{matrix} ∄ lim_{x \to a^{-}} f (x) \lor ∄ lim_{x \to a^{+}} f (x) \\ Note: not exists or \pm \infty \end{matrix}

\begin{matrix} f (x) = \frac{(\frac{1}{x} - \frac{1}{x + 1})}{\frac{1}{x - 1} - \frac{1}{x}} \\ 0, \pm 1 are discontinuities \\ f (x) = \frac{\frac{x + 1 - x}{x (x + 1)}}{\frac{x - x + 1}{x (x - 1)}} = \frac{x (x - 1)}{x (x + 1)} \\ lim_{x \to 0} \frac{x (x - 1)}{x (x + 1)} = - 1 ⟹ Removable \\ lim_{x \to 1} \frac{x (x - 1)}{x (x + 1)} = 0 ⟹ Removable \\ lim_{x \to (- 1)^{-}} \frac{x \overset{\to - 2}{\overset{⏞}{(x - 1)}}}{x \underset{\to 0^{-}}{\underset{⏟}{(x + 1)}}} = \infty ⟹ Essential \end{matrix}