\documentclass[11pt]{article} %\usepackage[T2A]{fontenc} %\usepackage[utf8]{inputenc} %\usepackage[russian]{babel} \usepackage[x11names, svgnames, rgb]{xcolor} \usepackage{tikz} \usetikzlibrary{arrows,shapes} \usepackage{scrextend} \input{intro} \lhead{\color{gray} Шарафатдинов Камиль 192} \rhead{\color{gray} ДЗ к 07.12 (\texttt{sol1202})} \title{ДЗ на 07.12} \author{Шарафатдинов Камиль БПМИ-192} \date{билд: \today} % -- Here bet dragons -- \begin{document} \maketitle \question{17}{ \[ \lim_{x \to \infty} f(x) = 0, \qquad \lim_{x \to \infty} g(x) = 0, \qquad \] Доказать, что \[ \begin{cases} f(x), g(x) \text{ -- дифф. в окрестности } \infty \quad (1)\\ \exists \lim_{x \to +\infty} f(x) = 0\\ \exists \lim_{x \to +\infty} g(x) = 0\\ g'(x) \neq 0\\ \exists \frac{f'(x)}{g'(x)} \end{cases} \implies \lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \frac{f'(x)}{g'(x)} \] } \[ (1) \Rightarrow f\braced{\frac{1}{t}}, g\braced{\frac{1}{t}} \text{ -- дифф. в окрестности } 0 \] \[ \lim_{x \to \infty} \frac{f'(x)}{g'(x)} = \lim_{y \to 0} \frac{(f(\frac{1}{t}))'}{(g(\frac{1}{t}))'} = \lim_{y \to 0} \frac{-t^2}{-t^2}\frac{f'(\frac{1}{t})}{g'(\frac{1}{t})} \overset{\text{по правилу Лопиталя}}{=} \lim_{y \to 0} \frac{f(\frac{1}{t})}{g(\frac{1}{t})} = \lim_{x \to \infty} \frac{f(x)}{g(x)} \qed \] \question{21}{ \begin{enumerate} \item $ x^{2/3} + y^{2/3} = a^{2/3} $ \item $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $ \end{enumerate} } 1. \begin{gather*} x^{2/3} + y^{2/3} = a^{2/3}\\ \frac{2}{3}x^{-1/3} + y' \frac{2}{3} y^{-1/3} = 0\\ x^{-1/3} + y' y^{-1/3} = 0\\ y' = -\frac{y^{1/3}}{x^{1/3}} = -\sqrt[3]{\frac{y}{x}} \end{gather*} 2. \begin{gather*} \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\\ \frac{2x}{a^2} + \frac{2yy'}{b^2} = 0\\ \frac{yy'}{b^2} = -\frac{x}{a^2}\\ y' = -\frac{x}{y}\cdot\frac{b^2}{a^2} \end{gather*} \question{21}{ \[ r = ae^{m\varphi} \] } \begin{gather*} r = ae^{m\varphi}\\ \sqrt{x^2 + y^2} = ae^{m \arctan{\frac{y}{x}}}\\ \frac{2x + 2yy'}{2\sqrt{x^2 + y^2}} = ae^{m \arctan{\frac{y}{x}}} m (\arctan{\frac{y}{x}})' = ae^{m \arctan{\frac{y}{x}}} m \frac{1}{1 + \braced{\frac{y}{x}}^2} \frac{y'x - y}{x^2}\\ \frac{x + yy'}{y'x - y} = mae^{m \varphi}\frac{1}{\frac{x^2 + y^2}{x^2}} \frac{1}{x^2} \sqrt{x^2 + y^2} = \frac{mae^{m \varphi}}{r} = \mu\\ \frac{x + yy'}{y'x - y} = \mu\\ x + yy' = \mu (y'x - y)\\ x + \mu y = \mu y'x - yy'\\ y' = \frac{x + \mu y}{\mu x - y} = \frac{x + \frac{mae^{m \varphi}}{r}y}{\frac{mae^{m \varphi}}{r}x - y} = \frac{xr + ymae^{m \varphi}}{xmae^{m \varphi} - yr} \end{gather*} \question{23}{ \[ y_1 = ax^2, \qquad\qquad y_2 = \ln x \] } Чтобы кривые касались, достаточно, чтобы их функции и производные были равны в некоторой точке: \[ \begin{cases} y_1 = y_2\\ y_1' = y_2'\\ \end{cases} \] \[ 2ax = \frac{1}{x} \implies x = \sqrt\frac{1}{2a} \] \[ ax^2 = \ln x \overset{\text{подставим $x$}}{\implies} \frac{1}{2} = \ln {\sqrt\frac{1}{2a}} \implies \frac{1}{2a} = e \implies a = \frac{1}{2e} \] \question{24}{ \[ x = \frac{2t + t^2}{1 + t^3}, \qquad y = \frac{2t - t^2}{1 + t^3} \] } $ \exists \varphi^{-1}(x): xt^3 - t^2 - 2t + x = 0 $ и функции дифференцируемы в окрестностях нужных точек, поэтому $f'(x_0) = \frac{x'(t_0)}{y'(t_0)}$ \[ f'(x) = \frac{\braced{\frac{2t - t^2}{1 + t^3}}'}{\braced{\frac{2t + t^2}{1 + t^3}}'} = \frac{(2 - 2t)(1 + t^3) - 3t^3(2t - t^2)} {(2 + 2t)(1 + t^3) - 3t^3(2t + t^2)} = \frac{2 - 2t + 2t^3 - 2t^4 - 6t^4 + 3t^5} {2 + 2t + 2t^3 + 2t^4 - 6t^4 - 3t^5} \] \begin{enumerate}[(a)] \item $t = 0: $ $f'(x) = 1$\\ \[ 1 \cdot(y - 0) + (x - 0) = 0 \Rightarrow y + x = 0 \] \item $t = 1: $ $f'(x) = 3$\\ \[ 3 \cdot(y - 1.5) + (x - 1.5) = 0 \Rightarrow 3y + x - 3 = 0 \] \item $t = +\infty: $ $f'(x) = -1$\\ \[ -1 \cdot(y - 0) + (x - 0) = 0 \Rightarrow -y + x = 0 \] \end{enumerate} \question{25.a}{ \[ \lim_{x \to 0} \frac{\ln \cos ax}{\ln \cos bx} = \frac{a^2}{b^2} \] } \begin{align*} &\lim_{x \to 0} \frac{\ln \cos ax}{\ln \cos bx} \overset{\text{Лопиталь}}= \lim_{x \to 0} \frac{-a \sin ax}{\cos ax} \frac{\cos bx}{-b \sin bx} = \lim_{x \to 0} \frac{\sin ax}{\sin bx} \cdot \frac{a\cos bx}{b\cos ax} =\\\\ &\lim_{x \to 0} \frac{a(bx) \sin ax}{(ax)b \sin bx} \cdot \frac{a\cos bx}{b\cos ax} \overset{\text{1й зам. предел}}= \lim_{x \to 0} \frac{a}{b} \cdot \frac{a\cos bx}{b\cos ax} = \frac{a^2}{b^2} \end{align*} \question{25.b}{ \[ \lim_{x \to a} \frac{a^x - x^a}{x - a} = a^a (\ln a - 1) \] } \[ \lim_{x \to a} \frac{a^x - x^a}{x - a} = \lim_{x \to a} \frac{e^{x\ln a} - x^a}{x - a} \overset{\text{Лопиталь}}= \lim_{x \to a} \frac{\ln a \cdot a^x - ax^{a - 1}}{1} \overset{\text{ф-я непрерывна}}= \frac{\ln a \cdot a^a - a \cdot a^{a - 1}}{1} = a^a (\ln a - 1) \] \question{25.c}{ \[ \lim_{x \to 1} \frac{1}{\ln x} - \frac{1}{x - 1} = \frac{1}{2} \] } \begin{align*} \lim_{x \to 1} \frac{1}{\ln x} - \frac{1}{x - 1} = \lim_{x \to 1} \frac{x - 1 - \ln x}{(x - 1) \ln x} \overset{\text{Лопиталь}}= \lim_{x \to 1} \frac{1 - \frac{1}{x}}{\ln x + \frac{x - 1}{x}} \overset{\cdot \frac{x}{x}}= \lim_{x \to 1} \frac{x - 1}{x \ln x + x - 1} \overset{\text{Лопиталь}}=\\ =\lim_{x \to 1} \frac{1}{1 + \ln x + 1} = \frac{1}{2} \end{align*} \question{25.c}{ \[ \frac{(1 + x)^\frac{1}{x} - e}{x} = -\frac{e}{2} \] } \begin{align*} &\lim_{x \to 0}\frac{(1 + x)^\frac{1}{x} - e}{x} = \lim_{x \to 0}\frac{e^\frac{\ln (x + 1)}{x} - e}{x} \overset{\text{Лопиталь}}=\\ = &\lim_{x \to 0}\frac{e^\frac{\ln (x + 1)}{x} \cdot \frac{\frac{x}{x + 1} - \ln(x + 1)}{x^2}}{1} = \lim_{x \to 0}e^\frac{\ln (x + 1)}{x} \cdot \frac{\frac{x}{x + 1} - \ln(x + 1)}{x^2} \overset{\text{св-ва пределов}}=\\ = &\lim_{x \to 0}e^\frac{\ln (x + 1)}{x} \cdot \lim_{x \to 0} \frac{\frac{x}{x + 1} - \ln(x + 1)}{x^2} = \lim_{x \to 0}(x + 1)^\frac{1}{x} \cdot \lim_{x \to 0} \frac{x - (x + 1)\ln(x + 1)}{(x + 1)x^2} =\\ = &e \lim_{x \to 0} \frac{x - (x + 1)\ln(x + 1)}{(x + 1)x^2} \overset{\text{св-ва пределов}}= e \lim_{x \to 0} \frac{1}{x + 1} \lim_{x \to 0} \frac{x - (x + 1)\ln(x + 1)}{x^2} =\\ = &e \lim_{x \to 0} \frac{x - x\ln(x + 1) - \ln(x + 1)}{x^2} \overset{\text{св-ва пределов}}= e \left( \lim_{x \to 0}\frac{x - \ln(x + 1)}{x^2} - \lim_{x \to 0}\frac{x\ln(x + 1)}{x^2} \right) \overset{\text{два Лопиталя}}=\\ = &e \left( \lim_{x \to 0} \frac{1 - \frac{1}{x + 1}}{2x} - \lim_{x \to 0} \frac{\frac{1}{x + 1}}{1} \right) = e \left( \lim_{x \to 0} \frac{x + 1 - 1}{2x(x + 1)} - 1 \right) = e \left( \lim_{x \to 0} \frac{1}{2(x + 1)} - 1 \right) = e \left( \frac{1}{2} - 1 \right) = -\frac{e}{2} \end{align*} \question{25.e}{ \[ \lim_{x \to 1} \frac{1}{\ln x} - \frac{1}{x - 1} = \frac{1}{2} \] } \begin{align*} &\lim_{x \to 1} \frac{1}{\ln x} - \frac{1}{x - 1}= \lim_{x \to 1} \frac{x - 1 - \ln x}{(x - 1)\ln x} \overset{\text{Лопиталь}}=\\ = &\lim_{x \to 1} \frac{1 - \frac{1}{x}}{\ln x + \frac{x - 1}{x}} = \lim_{x \to 1} \frac{x - 1}{x\ln x + x - 1} \overset{\text{Лопиталь}}=\\ = &\lim_{x \to 1} \frac{1}{1 + \ln x + 1} = \frac{1}{2} \end{align*} \end{document}