\ANG{
Berechnen Sie folgende Grenzwerte.
\begin{enumerate}
\item $\D\lim_{x \to 2}\frac{|x|+2}{|x-2|}$,
$\D\lim_{x \to 2}\frac{x-2}{|x-2|}$
\item $\D\lim_{x \to \infty}\frac{x-|x|}{1+x}$,
$\D\lim_{x \to -\infty}\frac{x-|x|}{1+x}$
\item $\D\lim_{x \to 0}\frac{1}{|2x|-4x^2}$,
$\D\lim_{x \to \frac{1}{2}}\frac{1}{|2x|-4x^2}$,
$\D\lim_{x \to -\frac{1}{2}}\frac{1}{|2x|-4x^2}$
\item $\D\lim_{x \to 0}\frac{x}{|x|+x^2}$
\end{enumerate}

}
\LSG{
\begin{enumerate}
\item $\infty$, $\D\lim_{x \to 2+0}\frac{x-2}{|x-2|}=1$,
$\D\lim_{x \to 2-0}\frac{x-2}{|x-2|}=-1$
\item $0$, $2$
\item $\D\lim_{x \to 0\pm 0}\frac{1}{|2x|-4x^2}=\infty$,
$\D\lim_{x \to \frac{1}{2}\pm 0}\frac{1}{|2x|-4x^2}=\mp\infty$,
$\D\lim_{x \to -\frac{1}{2}\pm 0}\frac{1}{|2x|-4x^2}=\pm\infty$
\item $\D\lim_{x \to 0\pm 0}\frac{x}{|x|+x^2}=\pm 1$
\end{enumerate}
}
