\documentclass[11pt]{article} %\usepackage[T2A]{fontenc} %\usepackage[utf8]{inputenc} %\usepackage[russian]{babel} \usepackage[sfdefault,condensed,scaled=0.8]{roboto} \usepackage{inconsolata} \setmonofont[Scale=0.85]{Inconsolata} \setlength\headheight{13.6pt} \usepackage{ amsmath, amsthm, amssymb, mathtools, graphicx, subfig, float, listings, xcolor, fancyhdr, sectsty, hyperref, enumerate, framed, comment } \usepackage[shortlabels]{enumitem} \flushbottom % Uncomment to make text fill the entire page \usepackage[bottom]{footmisc} % Anchor footnotes to bottom of page \renewcommand{\baselinestretch}{1.06} % Adjust line spacing %\setlength\parindent{0pt} % Remove paragraph indentation \usepackage{geometry}\geometry{letterpaper, % Set page margins left=1in, right=1in, top=0.8in, bottom=0.9in, headsep=.1in } \setlength\FrameSep{0.75em} \setlength\OuterFrameSep{\partopsep} \newenvironment{cframed}[1][gray] {\def\FrameCommand{\fboxsep=\FrameSep\fcolorbox{#1}{white}}% \MakeFramed {\advance\hsize-\width \FrameRestore}} {\endMakeFramed} \newcommand{\question}[2]{\doubleskip\begin{cframed}\noindent \textbf{#1} #2\end{cframed}} \newcommand{\withbraces}[1]{\left( #1 \right)} \DeclarePairedDelimiter\ceil{\lceil}{\rceil} \DeclarePairedDelimiter\floor{\lfloor}{\rfloor} \DeclareMathOperator{\tg}{tg} \DeclareMathOperator{\ctg}{ctg} \newcommand{\sinx}{\sin x} \newcommand{\cosx}{\cos x} \newcommand{\tgx}{\tg x} \newcommand{\doubleskip}{\bigskip \bigskip} \newcommand{\osmall}[1]{\overline{o}\left( #1 \right)} % -- Flush left for 'enumerate' numbers %\setlist[enumerate]{wide=0pt, leftmargin=21pt, labelwidth=0pt, align=left} \hypersetup{colorlinks=true, linkcolor=magenta} % -- Left/right header text and footer (to appear on every page) -- \pagestyle{fancy} \renewcommand{\footrulewidth}{0.4pt} \renewcommand{\headrulewidth}{0.4pt} \lhead{\color{gray} \texttt{sol1028}} \rhead{\color{gray} Шарафатдинов Камиль БПМИ192} \cfoot{} \rfoot{\thepage} % -- Here bet dragons -- \begin{document} Здесь \textbf{не} записано: 19bcd, 20b \question{9.a}{ \[ \lim_{x\to\pi} \frac{\sin{mx}}{\sin{nx}} = (-1)^{m + n} \cdot \frac{m}{n} \] } Пусть $y = \pi - x$ или $x = \pi - y$ \begin{flalign*} &\lim_{x \to \pi} \frac{\sin{mx}}{\sin{nx}} = \lim_{y \to 0} \frac{\sin(m\pi - my)}{\sin(n\pi - ny)} = \lim_{y \to 0} \frac{\sin(m\pi)\cos(my) - \sin(my)\cos(m\pi)} {\sin(n\pi)\cos(ny) - \sin(ny)\cos(n\pi)} = \\ = &\lim_{y \to 0} \frac{\sin(my)\cos(m\pi)}{\sin(ny)\cos(n\pi)} = \lim_{y \to 0} (-1)^{m + n} \cdot \frac{\sin(my)}{\sin(ny)} = (-1)^{m + n} \cdot \frac{m}{n} \end{flalign*} \question{9.b}{ \[ \lim_{x \to 0} \frac{\tg{x}}{x} = 1 \] } \[ \lim_{x \to 0} \frac{\tg{x}}{x} = \lim_{x \to 0} \frac{\sinx}{x\cosx} = \lim_{x \to 0} 1 \cdot \frac{1}{\cosx} = 1 \] \question{9.c}{ \[ \lim_{x \to 0} x \cdot \sin{\frac{1}{x}} = 0 \] } Так как $x \to 0$, а $\sin{\frac{1}{x}}$ -- ограничен, то $x\sin{\frac{1}{x}}$ стремится к 0. \question{9.d}{ \[ \lim_{x \to \infty} \frac {\withbraces{x - \sqrt{x^2 - 1}} ^ n + \withbraces{x + \sqrt{x^2 - 1}} ^ n} {x^n} = 2^n \] } \doubleskip Лемма 1: $\displaystyle \lim_{x \to \infty} \frac{\withbraces{x - \sqrt{x^2 - 1}} ^ n}{x^n} = 0$. \[ \lim_{x \to \infty} \withbraces{\frac{x - \sqrt{x^2 - 1}}{x}}^n = \lim_{x \to \infty} \withbraces{1 - \frac{\sqrt{x^2 - 1}}{x}}^n = \lim_{x \to \infty} \withbraces{1 - \sqrt{1 - \frac{1}{x^2}}}^n = (1 - 1)^n = 0 \] \doubleskip Лемма 2: $\displaystyle \lim_{x \to \infty} \frac{\withbraces{x + \sqrt{x^2 - 1}} ^ n}{x^n} = 2^n$. \[ \lim_{x \to \infty} \withbraces{\frac{x + \sqrt{x^2 - 1}}{x}}^n = \lim_{x \to \infty} \withbraces{1 + \frac{\sqrt{x^2 - 1}}{x}}^n = \lim_{x \to \infty} \withbraces{1 + \sqrt{1 - \frac{1}{x^2}}}^n = (1 + 1)^n = 2^n \] \doubleskip По свойству пределов (предел суммы - сумма пределов, если они существуют) и по леммам: \[ \lim_{x \to \infty} \frac {\withbraces{x - \sqrt{x^2 - 1}} ^ n + \withbraces{x + \sqrt{x^2 - 1}} ^ n} {x^n} = 0 + 2^n = 2^n \] \question{9.e}{ \[ \lim_{x \to 1} \frac{\withbraces{1 - \sqrt{x}} \withbraces{1 - \sqrt[3]{x}} \cdot \ldots \cdot \withbraces{1 - \sqrt[n]{x}}} {\withbraces{1 - x}^{n - 1}} = \frac{1}{n!} \] } Заметим, что \[ t_k = 1 - x = \withbraces{1 - \sqrt[k]{x}} \withbraces{1 + \sqrt[k]{x} + \sqrt[k]{x}^2 + \ldots + \sqrt[k]{x}^{k - 1}} \] А если $x \to 1$, то в пределе $\displaystyle \lim_{x \to 1} t_k = \withbraces{1 - \sqrt[k]{x}} \cdot k$ Тогда \begin{align*} &\lim_{x \to 1} \frac{\withbraces{1 - \sqrt{x}} \withbraces{1 - \sqrt[3]{x}} \cdot \ldots \cdot \withbraces{1 - \sqrt[n]{x}}} {\withbraces{1 - x}^{n - 1}} =\\ = &\lim_{x \to 1} \frac{1}{n!} \cdot \frac{\withbraces{1 - \sqrt{x}} \withbraces{1 - \sqrt[3]{x}} \cdot \ldots \cdot \withbraces{1 - \sqrt[n]{x}}} {\withbraces{1 - \sqrt{x}} \withbraces{1 - \sqrt[3]{x}} \cdot \ldots \cdot \withbraces{1 - \sqrt[n]{x}}} = \frac{1}{n!} \end{align*} \begin{comment} \question{10.a}{ \[ \lim_{x \to \infty} \left( \sqrt[n]{(x - a_1)(x - a_2) \cdot \ldots \cdot (x - a_n)} - x \right) = 0 \] } Пусть $A_k$ - сумма всевозможных произведений $a_i$, из $k$ членов: \begin{flalign*} A_1 &= a_1 + a_2 + \ldots + a_n\\ A_2 &= a_1a_2 + a_1a_3 + \ldots + a_{n - 1}a_n\\ \vdots \ \ &\\ A_n &= a_1a_2\ldots a_n \end{flalign*} Тогда \begin{flalign*} &\lim_{x \to \infty} \left( \sqrt[n]{(x - a_1)(x - a_2) \cdot \ldots \cdot (x - a_n)} - x \right) =\\ &\lim_{x \to \infty} x \left( \sqrt[n]{\frac{(x - a_1)(x - a_2) \cdot \ldots \cdot (x - a_n)}{x^n}} - 1 \right) =\\ &\lim_{x \to \infty} x \left( \sqrt[n]{\frac{x^n - A_1 x^{n - 1} + A_2 x^{n - 2} - \ldots + (-1)^n A_n}{x^n}} - 1 \right) =\\ &\lim_{x \to \infty} x \left( \sqrt[n]{\frac{x^n}{x^n} - \frac{A_1 x^{n - 1}}{x^n} + \ldots + \frac{(-1)^n A_n}{x^n}} - 1 \right) =\\ &\lim_{x \to \infty} x \left( \sqrt[n]{1 - o(x)} - 1 \right) =\\ &\lim_{x \to \infty} \left( x\sqrt[n]{1 - o(x)} - x \right) = 0 \end{flalign*} \question{10.b}{ \[ \lim_{x \to \infty} \frac{1 - \cosx \cos 2x \cos 3x}{1 - \cosx} \] } \[ \cosx \cos 2x \cos 3x = \cosx (2\cos^2 x - 1) (4\cos^3 x - 3\cosx) \] \[ \frac{1 - \cosx (2\cos^2 x - 1) (4\cos^3 x - 3\cosx)}{1 - \cosx} = \begin{cases} 1, \text{если } x = \frac{\pi}{2} + 2 \pi n\\ \frac{3}{2}, \text{если } x = \frac{\pi}{3} + 2 \pi n \end{cases} \] Значит, предела не существует. \end{comment} \question{14}{ \[ \lim_{x \to a} \withbraces{\frac{1 + x}{2 + x}}^{\frac{1 - \sqrt{x}}{1 - x}} \] } Если $a = 0$, то, очевидно, предел равен $\frac{1}{2}$ Если $a = 1$: \[ \lim_{x \to 1} \withbraces{\frac{1 + x}{2 + x}}^{\frac{1 - \sqrt{x}}{1 - x}} = \lim_{x \to 1} \withbraces{\frac{1 + x}{2 + x}}^{\frac{1}{1 + \sqrt{x}}} = \withbraces{\frac{2}{3}}^{\frac{1}{2}} = \sqrt{\frac{2}{3}} \] Если $a = +\infty$: \begin{align*} &\lim_{x \to \infty} \withbraces{\frac{1 + x}{2 + x}}^{\frac{1 - \sqrt{x}}{1 - x}} = \lim_{x \to \infty} \withbraces{1 - \frac{1}{2 + x}}^{\frac{1}{1 + \sqrt{x}}} =\\ &\lim_{x \to \infty} \withbraces{1 - \frac{1}{2 + x}}^{\frac{1}{1 + \sqrt{x}}} = \lim_{x \to \infty} e^{- \frac{1}{2 + x} \frac{1}{1 + \sqrt{x}}} = e^0 = 1 \end{align*} \question{15.a}{ \[ \lim_{n \to \infty} \left(\cos{\frac{x}{\sqrt{n}}}\right)^n = ?? \] } \[ \lim_{n \to \infty} \left(\cos{\frac{x}{\sqrt{n}}}\right)^n = \lim_{n \to \infty} \left(1 - \frac{x^2}{2n} + \overline{o}\left(\frac{x^2}{n}\right)\right)^n = e^{-\frac{x^2}{2}} \] \question{15.b}{ \[ \lim_{x \to 0} \sqrt[x]{1 - 2x} = \frac{1}{e^2} \] } $y = 1/x$ \[ \lim_{x \to 0} \sqrt[x]{1 - 2x} = \lim_{y \to \infty} \withbraces{1 - \frac{2}{y}}^y = e^{-2} = \frac{1}{e^2} \] \question{15.c}{ \[ \lim_{x \to \frac{\pi}{4}} \left(\tg x\right)^{\tg 2x} = e^{-1} \] } Пусть $y + 1 = \tg x$, \qquad $\displaystyle z = \frac{1}{y} = \frac{1}{\tg x - 1}$, \qquad $y \to 0$, $z \to \infty$. \begin{align*} &\lim_{x \to \frac{\pi}{4}} \left(\tg x\right)^{\tg 2x} = \lim_{x \to \frac{\pi}{4}} \left(1 + y\right)^{\frac{2\tg x}{1 - \tg^2 x}} = \lim_{x \to \frac{\pi}{4}} \left(1 + y\right)^{\frac{2(y + 1)}{1 - (y + 1)^2}} = \lim_{x \to \frac{\pi}{4}} \left(1 + y\right)^{-\frac{2(y + 1)}{y(y + 2)}} =\\ = &\lim_{x \to \frac{\pi}{4}} \left(\left(1 + \frac{1}{z}\right)^z\right)^{-\frac{2(y + 1)}{(y + 2)}} = \lim_{x \to \frac{\pi}{4}} \exp\left(-\frac{2(y + 1)}{y + 2}\right) = \lim_{x \to \frac{\pi}{4}} \exp{\left(-\frac{2\osmall{1} + 2}{\osmall{1} + 2}\right)} = e^{-1} \end{align*} \question{15.d}{ \[ \lim_{x \to 0} \frac{\sqrt{1 + \tg x} - \sqrt{1 + \sin x}}{x^3} = \frac{1}{4} \] } \begin{align*} &\lim_{x \to 0} \frac{\sqrt{1 + \tg x} - \sqrt{1 + \sin x}}{x^3} = \lim_{x \to 0} \frac{1 + \tg x - 1 - \sin x}{x^3 \left(\sqrt{1 + \tg x} + \sqrt{1 + \sin x}\right)} =\\ = &\lim_{x \to 0} \frac{\sin x \left( \frac{1}{\cosx} - 1 \right)}{x^3 \left(\sqrt{1 + \tg x} + \sqrt{1 + \sin x}\right)} = \lim_{x \to 0} \frac{\sinx}{x} \frac{1 - \cosx}{x^2 \cosx \left(\sqrt{1 + \tg x} + \sqrt{1 + \sin x}\right)} =\\ = &\lim_{x \to 0} \frac{\sinx}{x} \frac{1 - \left(1 - \frac{x^2}{2} + \overline{o}\withbraces{x^2}\right)}{x^2 \cosx \left(\sqrt{1 + \tg x} + \sqrt{1 + \sin x}\right)} = \lim_{x \to 0} \frac{\sinx}{x} \frac{x^2 + 2\overline{o}\withbraces{x^2}}{2x^2 \cosx \left(\sqrt{1 + \tg x} + \sqrt{1 + \sin x}\right)} =\\ = &1 \cdot \frac{1}{2 \cdot 1 \cdot (1 + 1)} = \frac{1}{4} \end{align*} \question{15.e}{ \[ \lim_{x \to \infty} \sin \sqrt{x + 1} - \sin \sqrt{x} = 0 \] } \[ \sin \sqrt{x + 1} - \sin \sqrt{x} = 2\cos \frac{\sqrt{x + 1} + \sqrt{x}}{2} \sin \frac{\sqrt{x + 1} - \sqrt{x}}{2} \] \[ \sqrt{x + 1} - \sqrt{x} = (\sqrt{x + 1} - \sqrt{x}) \frac{\sqrt{x + 1} + \sqrt{x}}{\sqrt{x + 1} + \sqrt{x}} = \frac{x + 1 - x}{\sqrt{x + 1} + \sqrt{x}} = \frac{1}{\sqrt{x + 1} + \sqrt{x}} \] \[ \lim_{x \to \infty} \sin \frac{\sqrt{x + 1} - \sqrt{x}}{2} = \lim_{x \to \infty} \sin \frac{\frac{1}{\sqrt{x + 1} + \sqrt{x}}}{2} = \sin\frac{0}{2} = 0 \] Так как $\cos \frac{\sqrt{x + 1} + \sqrt{x}}{2}$ ограничен, а $\sin \frac{\sqrt{x + 1} - \sqrt{x}}{2}$ стремится к $0$, то их произведение также стремится к $0$. \question{15.f}{ \[ \lim_{x \to \infty} x \left(\ln (x + 1) - \ln x\right) = 1 \] } \[ \lim_{x \to \infty} x \left(\ln (x + 1) - \ln x\right) = \lim_{x \to \infty} x \ln \frac{x + 1}{x} = \lim_{x \to \infty} \ln \left(\left(1 + \frac{1}{x}\right)^x\right) = \ln e = 1 \] \question{16.a}{ \[ \lim_{x \to \infty} \frac{x^n}{a^x} = 0, \qquad \text{где } a > 1, n \in \mathbb{N} \] } \[ \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n = \lim_{n \to \infty} \left(1 + \frac{1}{\frac{n}{x}}\right)^n = \lim_{n \to \infty} \left(1 + \frac{1}{\frac{n}{x}}\right)^{n \frac{x}{n} \frac{n}{x}} = \lim_{n \to \infty} \left(1 + \frac{1}{\frac{n}{x}}\right)^{\frac{n}{x} x} = e^x \] По неравенству Бернулли: $\displaystyle \left(1 + \frac{x}{n}\right)^n \geq 1 + n \frac{x}{n} = 1 + x$. Поэтому $e^x \geq 1 + x$ \begin{align*} &\lim_{x \to \infty} \frac{x^n}{\left(e^{\ln a}\right)^x} = \lim_{x \to \infty} \frac{x^n}{\left(e^{\frac{x \ln a}{2n}}\right)^{2n}} = \lim_{x \to \infty} \frac{x^n}{\left(1 + \frac{x \ln a}{2n}\right)^{2n}} = \lim_{x \to \infty} \left(\frac{x}{\left(1 + \frac{x \ln a}{2n}\right)^{2}}\right)^n =\\ &\lim_{x \to \infty} \left(\frac{x}{1 + \frac{x \ln a}{n} + \frac{(x \ln a)^2}{4n^2}}\right)^n = \lim_{x \to \infty} \left(\frac{1}{\frac{1}{x} + \frac{\ln a}{n} + \frac{x \ln^2 a}{4n^2}}\right)^n = 0 \end{align*} \question{16.b}{ \[ \lim_{x \to \infty} \frac{\log_a x}{x^\varepsilon} = 0, \qquad a, \varepsilon > 0, a \neq 1 \] } Пусть $y = \ln x$ \[ \lim_{x \to \infty} \frac{\log_a x}{x^\varepsilon} = \frac{1}{\ln a}\ \ \lim_{x \to \infty} \frac{\ln x}{x^\varepsilon} = \frac{1}{\ln a}\ \ \lim_{y \to \infty} \frac{y}{e^{\ln \left(x^\varepsilon\right)}} = \frac{1}{\ln a}\ \ \lim_{y \to \infty} \frac{y}{e^{by}} \] По \textbf{16.a}: $\displaystyle \lim_{x \to \infty} \frac{x}{e^{bx}} = 0$, поэтому \[ \frac{1}{\ln a}\ \ \lim_{y \to \infty} \frac{y}{e^{by}} = \frac{1}{\ln a} \cdot 0 = 0 \] \question{17.a}{ \[ \lim_{x \to 0} x \ln x = 0 \] } Пусть $y = \frac{1}{x}$ \[ \lim_{x \to 0} x \ln x = \lim_{y \to \infty} \frac{\ln{\frac{1}{y}}}{y} = \lim_{y \to \infty} \frac{\ln 1 - \ln y}{y} = \lim_{y \to \infty} -\frac{\ln y}{y} \quad \overset{\text{по \textbf{16.b}}}{=} \quad -0 = 0 \] \question{17.b}{ \[ \lim_{x \to 1} (1 - x) \log_x 2 = -\ln 2 \] } \[ \lim_{x \to 1} (1 - x) \log_x 2 = \lim_{x \to 1} (1 - x) \frac{\ln 2}{\ln x} = \ln 2 \lim_{x \to 1} \frac{1 - x}{\ln x} \] Пусть $t = 1 - x$. \qquad $t \to 0$ \[ \ln 2 \lim_{x \to 1} \frac{1 - x}{\ln x} = \ln 2 \lim_{t \to 0} \frac{t}{\ln (1 - t)} = \ln 2 \lim_{t \to 0} \frac{1}{\frac{1}{t} \ln (1 - t)} = \ln 2 \lim_{t \to 0} \frac{1}{\ln (1 - t)^{\frac{1}{t}}} \] Пусть $u = t^{-1}$. \qquad $u \to \infty$ \[ \ln 2 \lim_{t \to 0} \frac{1}{\ln (1 - t)^{\frac{1}{t}}} = \ln 2 \lim_{u \to \infty} \frac{1}{\ln (1 - \frac{1}{u})^u} = \ln (2) \cdot \frac{1}{-1} = -\ln 2 \] \question{19.a}{ \[ \lim_{x \to 0} \frac{\tg x - \sin x}{\sin^3 x} = \frac{1}{2} \] } \begin{align*} &\lim_{x \to 0} \frac{\tg x - \sin x}{\sin^3 x} = \lim_{x \to 0} \frac{\sin x \left(\frac{1}{\cos x} - 1\right)}{\sin^3 x} = \lim_{x \to 0} \frac{\frac{1 - \cos x}{\cos x}}{\sin^2 x} =\\\\ = &\lim_{x \to 0} \frac{1 - \cos x}{\cos x (1 - \cos^2 x)} = \lim_{x \to 0} \frac{1}{\cos x (1 + \cos x)} = \frac{1}{1 \cdot 2} = \frac{1}{2} \end{align*} \question{20.a}{ \[ \lim_{x \to 0} \frac{\sqrt{1 - \cos x^2}}{1 - \cos x} = \sqrt{2} \] } \begin{align*} &\lim_{x \to 0} \frac{\sqrt{1 - \cos x^2}}{1 - \cos x} = \lim_{x \to 0} \frac{\sqrt{1 - (1 - \frac{x^4}{2} + \osmall{x^4})}}{1 - (1 - \frac{x^2}{2} + \osmall{x^2})} = \lim_{x \to 0} \frac{\sqrt{\frac{x^4}{2} - \osmall{x^4}}}{\frac{x^2}{2} - \osmall{x^2}} =\\ = &\lim_{x \to 0} \sqrt{ \frac{\frac{x^4}{2} - \osmall{x^4}}{\frac{x^4}{4} - x^2\osmall{x^2} + \osmall{x^4}} } = \lim_{x \to 0} \sqrt{ \frac{\frac{x^4}{2}}{\frac{x^4}{4} - x^2\osmall{x^2} + \osmall{x^4}} } = \sqrt{2} \end{align*} %\question{20.b}{ % \[ % \lim_{x \to 0} \frac{1 - \sqrt{\cos x}}{1 - \cos{\sqrt{x}}} = ?? % \] %} % \[ % \lim_{x \to 0} \frac{1 - \sqrt{\cos x}}{1 - \cos{\sqrt{x}}} = % \lim_{x \to 0} \frac{1 - \sqrt{1 - \frac{x^2}{2} + \osmall{x^2}}}{1 - \left(1 - \frac{x}{2} + \osmall{x}\right)} = % \] \question{21.b}{ \[ \lim_{x \to \infty} \left(\frac{2x^2 - x + 1}{2x^2 + x + 1}\right)^{\frac{x^2}{1 - x}} = e \] } \begin{align*} &\lim_{x \to \infty} \left(\frac{2x^2 - x + 1}{2x^2 + x + 1}\right)^{\frac{x^2}{1 - x}} = \lim_{x \to \infty} \left(1 - \frac{2x}{2x^2 + x + 1}\right)^{-\frac{x^2}{x - 1}} = \lim_{x \to \infty} \left(1 - \frac{1}{x + \frac{1}{2} + \frac{1}{2x}}\right)^{-\frac{x^2}{x - 1}} =\\ &\lim_{x \to \infty} \left(1 - \frac{1}{x + \frac{1}{2} + \frac{1}{2x}}\right)^ {-\frac{x + \frac{1}{2} + \frac{1}{2x}}{x + \frac{1}{2} + \frac{1}{2x}}\frac{x^2}{x - 1}} = \lim_{x \to \infty} e^{\frac{1}{x + \frac{1}{2} + \frac{1}{2x}}\frac{x^2}{x - 1}} = \lim_{x \to \infty} \exp\left({\frac{1}{x + \frac{1}{2} + \frac{1}{2x}}\frac{x^2}{x - 1}}\right) =\\ &\lim_{x \to \infty} \exp\left({\frac{x^2}{\left(x + \frac{1}{2} + \frac{1}{2x}\right)(x - 1)}}\right) = \lim_{x \to \infty} \exp\left({\frac{x^2}{x^2 - \frac{x}{2} - \frac{1}{2x}}}\right) = e \end{align*} \end{document}