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\rhead{\color{gray} Шарафатдинов Камиль БПМИ192}
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% -- 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}
