HW10

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\begin{document}
    \begin{tabular}{p{0.5\textwidth} p{0.5\textwidth}}
        \begin{flushleft}\bf{ПМИ-203-2 Егоров Егор}\end{flushleft} & \begin{flushright}27/11/2020\end{flushright}
    \end{tabular}


    \solution{9g}
    {
    \[
    \lim_{x \to 0} \dfrac{ \cos(a+2x) - 2\cos(a+x) + \cos(a) }  {x^2}
    \]

    \begin{flalign*}
        &\lim_{x \to 0} \dfrac{ \cos(a+2x) - 2\cos(a+x) + \cos(a) }  {x^2} =&\\=
        %
        &\lim_{x \to 0} \dfrac{ \cos a \cdot \cos 2x - \sin a \cdot \sin 2x - 2\cos a \cdot \cos x + 2\sin a \cdot \sin x + \cos a }  {x^2} =&\\=
        %
        &\lim_{x \to 0} \dfrac{ \cos a (\cos 2x - 2\cos x + 1) + 2 \sin a \cdot \sin x - 2\sin a \cdot \sin x \cdot \cos x }  {x^2} =&\\=
        %
        &\lim_{x \to 0} \dfrac{ \cos a (\cos^2 x - \sin^2 x - 2\cos x + 1) + 2 \sin a \cdot \sin x (1 - \cos x) }  {x^2} =&\\=
        %
        &\lim_{x \to 0} \dfrac{ \cos a (2\cos^2 x - 2\cos x) + 2 \sin a \cdot \sin x (1 - \cos x) }  {x^2} =&\\=
        %
        &\lim_{x \to 0} \dfrac{ 2 \cos a \cdot \cos x(\cos x - 1) + 2 \sin a \cdot \sin x (1 - \cos x) }  {x^2} =&\\=
        %
        &\lim_{x \to 0} \left (\dfrac {2 \cos a \cdot \cos x(\cos x - 1)}{x^2} +
        \dfrac{2 \sin a \cdot \sin x (1 - \cos x)}{x^2}
        \right) =&\\=
        %
        &\lim_{x \to 0}  \left(
        2 \cos a \cdot \cos x \cdot \left( -\dfrac{1}{2} \right) +
        2 \sin a \cdot \sin x \cdot \left(\dfrac{1}{2} \right)
        \right) =
        %
        \lim_{x \to 0} \left(
        2 \cos a \cdot \cos x \cdot \left( -\dfrac{1}{2} \right)
        \right) =&\\=
        %
        &- \cos a \cdot 1 = -\cos a
    \end{flalign*}

    \ans $-\cos a$
    }

    \newpage

    \solution{10c}
    {
    \[
    \lim_{x \to a} \dfrac{a^x - x^a}{x-a}, a > 0
    \]

    \begin{gather*}
        \lim_{x \to a} \dfrac{a^x - x^a}{x-a} =
        %
        \left \{
        \begin{gathered}
            y = x-a\\
            y \to 0
        \end{gathered}
        \right \} =
        %
        \lim_{y \to 0} \dfrac{a^{y+a} - (y+a)^a}{y} =
        %
        \lim_{y \to 0} \dfrac{a^a \left(a^y - \left(\dfrac{y+a}{a} \right)^a\right)} {x-a} =\\=
        %
        \lim_{y \to 0} \dfrac{a^a \left(a^y - \left(1 + \dfrac{y}{a} \right)^a\right)} {y} =
        %
        a^a \cdot \lim_{y \to 0}  \left(
            \dfrac{a^y-1}{y} -
            \dfrac{ \left(1 + \dfrac{y}{a} \right)^a - 1}{y}
        \right) =\\=
        %
        a^a \cdot \lim_{y \to 0}  \left(
        \dfrac{e^{y \ln a}-1}{y} -
        \dfrac{1}{a} \cdot \dfrac{ \left(1 + \dfrac{y}{a} \right)^a - 1}{\dfrac{y}{a}}
        \right) =
        %
        a^a \left( \lim_{y \to 0}  \left(
        \ln a \cdot \dfrac{e^{y \ln a}-1}{\ln a \cdot y}
        \right) -
        \dfrac{1}{a} \cdot a
        \right) =\\=
        %
        a^a \left(\ln a -
        \dfrac{1}{a} \cdot a
        \right) =
        %
        a^a(\ln a - 1)
    \end{gather*}

    \ans $a^a(\ln a - 1)$
    }

    \solution{10d}
    {
    \[
    \lim_{x \to a} \dfrac{\ln x - \ln a}{x - a}
    \]

    \begin{gather*}
        \lim_{x \to a} \dfrac{\ln x - \ln a}{x - a} =
        %
        \left \{ \begin{gathered}
            y= x - a\\
            y \to 0
        \end{gathered}
        \right\} =
        %
        \lim_{y \to 0} \dfrac{ \ln (y+a) - \ln a}{y} =
        %
        \lim_{y \to 0} \dfrac{ \ln \left(\dfrac{y + a}{a} \right) }{y} =
        %
        \lim_{y \to 0} \dfrac{ \ln \left(\dfrac{y}{a} + 1\right) }{y} =\\=
        %
        \lim_{y \to 0} \left( \dfrac{1}{a} \cdot \dfrac{ \ln \left(\dfrac{y}{a} + 1\right) }{\dfrac{y}{a}} \right) =
        %
        \dfrac{1}{a} \cdot 1 = \dfrac{1}{a}
    \end{gather*}


    \ans $\dfrac{1}{a}$
    }

    \newpage

    \solution{10e}
    {
    \[
    \lim_{x \to 0} \dfrac{ \ln (x^2 + e^x) } { \ln (x^4 + e^{2x})}
    \]

    \begin{flalign*}
        &\lim_{x \to 0} \dfrac{ \ln (x^2 + e^x) } { \ln (x^4 + e^{2x})} =
        %
        \lim_{x \to 0} \left(
        \dfrac{ \ln \left(e^x\left( \dfrac{x^2}{e^x} + 1\right) \right) }
        { \ln \left(  e^{2x} \left(  \dfrac{x^4}{e^{2x}} + 1 \right) \right) }
        \right) =
        %
        \lim_{x \to 0} \left(
        \dfrac{x + \ln \left( \dfrac{x^2}{e^x} + 1 \right)}
        {2x + \ln \left( \dfrac{x^4}{e^{2x}} + 1 \right)}
        \right) =\\=
        %
        &\lim_{x \to 0} \left(
        \dfrac{x}{2x + \ln \left( \dfrac{x^4}{e^{2x}} + 1 \right)}
        +
        \dfrac{ \dfrac{x^2}{e^x} \left( \ln \left( \dfrac{x^2}{e^x} + 1 \right) \right) }
        {\dfrac{x^2}{e^x} \left( 2x + \ln \left( \dfrac{x^4}{e^{2x}} + 1 \right) \right)}
        \right) =\\=
        %
        &\lim_{x \to 0} \left(
        \dfrac{x}{2x + \ln \left( \dfrac{x^4}{e^{2x}} + 1 \right)}
        +
        \dfrac{ \dfrac{x^2}{e^x} }
        {2x + \ln \left( \dfrac{x^4}{e^{2x}} + 1 \right) }
        \right) =
        %
        \lim_{x \to 0} \left(
        \dfrac{x + \dfrac{x^2}{e^x}}{2x + \ln \left( \dfrac{x^4}{e^{2x}} + 1 \right)}
        \right) =\\=
        %
        &\lim_{x \to 0} \left(
        \dfrac{
            \dfrac{x +  x^2 / e^x}{x^4 / e^{2x}}
        }
        {
            \dfrac{2x + \ln \left( x^4 / e^{2x} + 1 \right)}
            {x^4 / e^{2x}}
        }
        \right) =
        %
        \lim_{x \to 0} \left(
        \dfrac{e^{2x} \left( x + \dfrac{x^2}{e^x} \right)}
        {x^4 \cdot \left( \dfrac{2x}{x^4 / e^{2x}} + 1 \right)}
        \right) =
        %
        \lim_{x \to 0} \left(
        \dfrac{e^{2x} x + e^x x^2}
        {x^4 \cdot \left( \dfrac{2x \cdot e^{2x}}{x^4} + 1 \right)}
        \right) =\\=
        %
        &\lim_{x \to 0} \left(
        \dfrac{e^{2x} x + e^x x^2}
        {2x \cdot e^{2x} + x^4}
        \right) =
        %
        \lim_{x \to 0} \left(
        \dfrac{e^{2x} + e^x x}
        {2 \cdot e^{2x} + x^3}
        \right) =
        %
        \dfrac{e^0 + e^0 \cdot 0} {2 \cdot e^0 + 0} =
        %
        \dfrac{1}{2}
    \end{flalign*}

    \ans $\dfrac{1}{2}$
    }

    \solution{1d}
    {
    \[
    \obg(f) + \osm(f) = \obg(f)
    \]
    \begin{gather*}
        \lim_{x \to a} \dfrac{\osm(f(x))}{f(x)} = 0 \rarrow
        \exists \delta_1 > 0: \forall x \in B'_{\delta_1}(a) \cap D:
        \left| \dfrac{\osm(f(x))}{f(x)} \right| < 1 \rarrow
        |\osm(f(x))| < |f(x)| \\
        %
        \exists \delta_2 > 0, C_2: \forall x \in B'_{\delta_2} \cap D:
        \left| \obg(f(x)) \right| \leq C_2 \cdot |f(x)|\\
        %
        \delta = \min \{ \delta_1, \delta_2 \}\\
        %
        \forall x \in B'_{\delta}(a) \cap D:
        | \obg(f(x)) + \osm(f(x)) | \leq
        | \obg(f(x)) | + | \osm(f(x))| \leq
        C_2 \cdot |f(x)| + |f(x)| = (C_2 + 1) |f(x)| \rarrow\\ \rarrow
        %
        \obg(f) + \osm(f) = \obg(f)
    \end{gather*}
    }

    \newpage
    \solution{1e}
    {
    \[
    \osm(f) = \obg(f)
    \]
    \begin{gather*}
        \lim_{x \to a} \dfrac{\osm(f(x))}{f(x)} = 0 \rarrow
        %
        \exists \delta > 0: \forall x \in B'_{\delta}(a) \cap D:
        \left| \dfrac{\osm(f(x))}{f(x)}  \right| < 1 \rarrow
        |\osm(f(x))| < |f(x)| = 1 \cdot |f(x)| \rarrow\\ \rarrow
        %
        \osm(f) = \obg(f)
    \end{gather*}
    }

    \solution{2e}
    {
    \[
    (x + x^2 + \osm(x^2))^2 = x^2 + \osm(x^3), \; x \to 0
    \]
    \begin{gather*}
        (x + x^2 + \osm(x^2))^2 = x^2 + \osm(x^3) \lrarrow\\ \lrarrow
        %
        x^2 + 2x(x^2 + \osm(x^2)) + (x^2 + \osm(x^2))^2 = x^2 + \osm(x^3) \lrarrow\\ \lrarrow
        %
        2x(x^2 + \osm(x^2)) + (x^2 + \osm(x^2))^2 = \osm(x^3)
    \end{gather*}
    \begin{gather*}
        \lim_{x \to 0} \left( \dfrac{2x(x^2 + \osm(x^2)) + (x^2 + \osm(x^2))^2} {x^3} \right) =
        %
        \lim_{x \to 0} \left( \dfrac{2(x^2 + \osm(x^2))}{x^2} +
        \dfrac{x^4 + 2x^2\osm(x^2) + (\osm(x^2))^2}{x^3} \right) =\\=
        %
        2 + \lim_{x \to 0} \dfrac{x^4 + 2x^4\osm(1) + x^4\osm(1)}{x^3} =
        %
        2 + \lim_{x \to 0} \left( x + \osm(x) \right) = 2
    \end{gather*}

    Значит, утверждение не справедливо.

    \ans Утверждение не справедливо
    }

    \solution{3b}
    {
    \[
    \lim_{x \to 0} \dfrac{(1+x)^x - 1} {1 - \cos x}
    \]
    \begin{gather*}
        \lim_{x \to 0} \dfrac{(1+x)^x - 1} {1 - \cos x} =
        %
        \lim_{x \to 0} \dfrac{e^{x \ln(1+x)} - 1} {\frac{1}{2}x^2 - \osm(x^2)} =
        %
        \lim_{x \to 0} \left(
        \dfrac{x\ln(1+x)}{x\ln(1+x)} \cdot
        \dfrac{e^{x \ln(1+x)} - 1} {\frac{1}{2}x^2 - \osm(x^2)}
        \right) =
        %
        \lim_{x \to 0} \dfrac{x\ln(1+x)} { \frac{1}{2} x^2 - \osm(x^2) } =\\=
        %
        \lim_{x \to 0} \dfrac{x(x + \osm(x))}{\frac{1}{2} x^2 - \osm(x^2) } =
        %
        \lim_{x \to 0} \dfrac{x^2 + x^2\osm(1)}{\frac{1}{2} x^2 - x^2\osm(1) } =
        %
        \lim_{x \to 0} \dfrac{1 + \osm(1)}{\frac{1}{2} - \osm(1) } =
        %
        \dfrac{1}{\frac{1}{2}} = 2
    \end{gather*}

    \ans 2
    }


    \vspace{10mm}
    \CheckedBox Увеличил шрифт с 10 до 12 pt

    \CheckedBox Где-то сделал выравнивание по левому краю - подумаю, как мне нравится больше \smiley

\end{document}