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\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}
