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%%%%%%%%%% Fill out the the definitions %%%%%%%%%
\def \name {Valentin Brandl}					%
\def \matrikel {108018274494}				%
\def \pname {Marvin Herrmann}				%
\def \pmatrikel {108018265436}				%
\def \qname {Pascal Brackmann}
\def \qmatrikel {108017113834}				%
\def \gruppe {2 (Mi 10-12 Andre)}
\def \uebung {3}								%
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\vspace*{-2cm}
\noindent \matrikel \quad \name \hfill \"Ubungsgruppe: \gruppe \\
\noindent \pmatrikel \quad \pname \\
\noindent \qmatrikel \quad \qname \\
\begin{center}{\Large \bf L\"osung f\"ur \"Ubung \# \uebung}\end{center}
}

\begin{document}
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\section*{Aufgabe 3.3}
\begin{enumerate}[1.]

	\item
		\begin{eqnarray*}
			\ln n &= o(e^{\sqrt{\ln n}}) \\
			\text{Anwendung des Satz von L'Hospital:} \\
			\lim\limits_{x \to x_0} \frac{f(x)}{g(x)} &= \lim\limits_{x \to x_0} \frac{f'(x)}{g'(x)} \\
			\Rightarrow &\lim\limits_{n \to \infty} \left( \frac{\frac{1}{n}}{e^{\sqrt{\ln(n)}}
				\frac{1}{2n\sqrt{\ln(n)}}} \right) \\
				&= \lim\limits_{n \to \infty} \left( \frac{\frac{1}{n}n}{e^{\sqrt{\ln(n)}}\frac{n}{2n\sqrt{\ln(n)}}}
			\right) \\
			&= \lim\limits_{n \to \infty} \left( \frac{1}{\frac{e^{\sqrt{\ln(n)}}}{2\sqrt{\ln(n)}}} \right) \\
			&= \lim\limits_{n \to \infty} \frac{1}{\infty} \\
		\end{eqnarray*}
		da $e^{\sqrt{\ln(n)}}$ schneller wächst als $2\sqrt{\ln(n)}$

		$\Rightarrow \ln n = o(e^{\sqrt{\ln n}})$

    \item
        \begin{eqnarray*}
            e^{\sqrt{\ln n}} &= o(n^{\varepsilon}) \text{ für alle } \varepsilon \in \mathbb{R}, \varepsilon > 0 \\
            \lim\limits_{n \to \infty} \frac{e^{\sqrt{\ln n}}}{e^{\ln(n^{\varepsilon})}} =
            \lim\limits_{n \to \infty} \frac{e^{\sqrt{\ln n}}}{e^{\varepsilon \cdot \ln n}} = 0
        \end{eqnarray*}

        Gilt für alle $\varepsilon > 0$, da $\ln n$ schneller wächst als $\sqrt{\ln n}$.

        $\Rightarrow e^{\sqrt{\ln n}} = o(n^{\varepsilon})$

\end{enumerate}

\end{document}

