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diff --git a/sol1028.tex b/sol1028.tex new file mode 100644 index 0000000..ddfd9df --- /dev/null +++ b/sol1028.tex @@ -0,0 +1,509 @@ +\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} |