Infi-1 13

Lagrange's mean value theorem

\begin{matrix} This is a generalization of Rolle’s theorem \\ Let f be continuous on [a, b] and differentiable on (a, b) \\ Then \exists ξ \in (a, b) : f^{'} (ξ) = \frac{f (b) - f (a)}{b - a} \end{matrix}

\begin{matrix} 0 < y \leq x < \frac{π}{2} \\ \frac{x - y}{\cos^{2} (y)} + \tan (y) \leq \tan x \leq \frac{x - y}{\cos^{2} (x)} + \tan y \\ Let x = y \\ \tan y \leq \tan x \leq \tan y ⟺ \tan x = \tan y \\ Let x \neq y \\ \frac{x - y}{\cos^{2} (y)} \leq \tan x - \tan y \leq \frac{x - y}{\cos^{2} (x)} \\ \frac{1}{\cos^{2} (y)} \leq \frac{\tan x - \tan y}{x - y} \leq \frac{1}{\cos^{2} (x)} \\ Let f (t) = \tan t on interval (y, x) \\ \tan t is continuous on (0, \frac{π}{2}) \supseteq [y, x] \\ \tan t is differentiable on (o, \frac{π}{2}) \supseteq (y, x) \\ ⟹ By the mean value theorem: \exists c \in (y, x) : f^{'} (c) = \frac{\tan x - \tan y}{x - y} \\ Let g (t) = f^{'} (t) = \frac{1}{\cos^{2} (t)} \\ \cos t is monotonically decreasing on (0, \frac{π}{2}) \\ ⟹ \cos^{2} (t) is monotonically decreasing on (0, \frac{π}{2}) \\ ⟹ \frac{1}{\cos^{2} (t)} is monotonically increasing on (o, \frac{π}{2}) \\ ⟹ \forall d \in (y, x) : \frac{1}{\cos^{2} (y)} \leq \frac{1}{\cos^{2} (d)} \leq \frac{1}{\cos^{2} (x)} \\ \exists c \in (y, x) : \frac{\tan x - \tan y}{x - y} = f^{'} (c) = g (c) = \frac{1}{\cos^{2} (c)} \\ ⟹ \frac{1}{\cos^{2} (y)} \leq \frac{\tan x - \tan y}{x - y} \leq \frac{1}{\cos^{2} (x)} \\ ⟹ \frac{x - y}{\cos^{2} (y)} + \tan (y) \leq \tan x \leq \frac{x - y}{\cos^{2} (x)} + \tan y \end{matrix}

\begin{matrix} Prove: \forall 0 < a < b : \frac{b - a}{b} < \ln (\frac{b}{a}) < \frac{b - a}{a} \\ Proof: \\ \ln (\frac{b}{a}) = \ln (b) - \ln (a) \\ b \neq a \\ \frac{b - a}{b} < \ln (\frac{b}{a}) < \frac{b - a}{a} \\ ⟺ \frac{1}{b} < \frac{\ln b - \ln a}{b - a} < \frac{1}{a} \\ Let f (t) = \ln t \\ \ln t is differentiable on (0, \infty) \supseteq [a, b] \\ By the mean value theorem: \exists c \in (a, b) : f^{'} (c) = \frac{\ln b - \ln a}{b - a} \\ Let g (t) = f^{'} (t) = \frac{1}{t} \\ \frac{1}{t} is monotonically decreasing on (0, \infty) \supseteq [a, b] \\ ⟹ \forall d \in (a, b) : \frac{1}{b} < \frac{1}{d} < \frac{1}{a} \\ \exists c \in (a, b) : \frac{\ln b - \ln a}{b - a} = f^{'} (c) = g (c) = \frac{1}{c} \\ ⟹ \frac{1}{b} < \frac{\ln b - \ln a}{b - a} < \frac{1}{a} \\ ⟹ \frac{b - a}{b} < \ln (\frac{b}{a}) < \frac{b - a}{a} \end{matrix}

\begin{matrix} Let f, g be continuous on [a, b] \\ and differentiable on (a, b) \\ Let \forall x \in (a, b) : g^{'} (x) \neq 0 \\ Then \exists c \in (a, b) : \frac{f^{'} (c)}{g^{'} (c)} = \frac{f (b) - f (a)}{g (b) - g (a)} \end{matrix}

\begin{matrix} Let 0 < x < 1 \\ Prove: \frac{\ln (x^{2} + 1)}{\tan x} < \frac{2 x}{x^{2} + 1} \\ Proof: \\ \frac{\ln (x^{2} + 1)}{\tan x} = \frac{\ln (x^{2} + 1) - \ln (0^{2} + 1)}{\tan x - \tan 0} \\ Let f (t) = \ln (t^{2} + 1) \\ Let g (t) = \tan t \\ f^{'} (t) = \frac{2 t}{t^{2} + 1} \\ \forall t \in (0, 1) : g^{'} (t) = \frac{1}{\cos^{2} (t)} \neq 0 \\ ⟹ By Cauchy’s theorem: \exists c \in (0, x) : \frac{f^{'} (c)}{g^{'} (c)} = \frac{\ln (x^{2} + 1) - \ln (0^{2} + 1)}{\tan x - \tan 0} \\ \frac{f^{'} (c)}{g^{'} (c)} = \frac{2 c}{c^{2} + 1} \cdot \cos^{2} (c) \\ \forall c \in (0, x) : \cos^{2} (c) \leq 1 ⟹ \frac{2 c}{c^{2} + 1} \cdot \cos^{2} (c) \leq \frac{2 c}{c^{2} + 1} \\ Let h (t) = \frac{2 t}{t^{2} + 1} \\ h^{'} (t) = \frac{2 (t^{2} + 1) - 4 t^{2}}{(t^{2} + 1)^{2}} = \frac{2 - 2 t^{2}}{(t^{2} + 1)^{2}} \\ t < 1 ⟹ t^{2} < 1 ⟹ 2 - 2 t^{2} > 0 ⟹ h^{'} (t) > 0 \\ ⟹ h (t) is monotonically increasing o n (0, 1) \\ ⟹ \forall d \in (0, x) : 0 < \frac{2 d}{d^{2} + 1} < \frac{2 x}{x^{2} + 1} \\ \exists c \in (0, x) : \frac{\ln (x^{2} + 1) - \ln (0^{2} + 1)}{\tan x - \tan 0} = \frac{f^{'} (c)}{g^{'} (c)} = \frac{2 c}{c^{2} + 1} < \frac{2 x}{x^{2} + 1} \\ ⟹ \frac{\ln (x^{2} + 1)}{\tan x} < \frac{2 x}{x^{2} + 1} \end{matrix}

\begin{matrix} lim_{x \to 1^{-}} (2 - x)^{\tan (\frac{π}{2} x)} = lim_{x \to 1^{-}} e^{\tan (\frac{π}{2} x) \ln (2 - x)} \\ lim_{x \to 1^{-}} \tan (\frac{π}{2} x) \ln (2 - x) = lim_{x \to 1^{-}} \frac{\ln (2 - x)}{{(\tan (\frac{π}{2} x))}^{- 1}} \\ \overset{L}{=} lim_{x \to 1^{-}} \frac{\frac{- 1}{2 - x}}{\frac{- \frac{π}{2}}{\sin^{2} (\frac{π}{2} x)}} = \frac{2}{π} \\ ⟹ lim_{x \to 1^{-}} (2 - x)^{\tan (\frac{π}{2} x)} = e^{2 / π} \end{matrix}