\ANG{
Berechnen Sie in nachvollziehbarer Weise die folgenden Grenzwerte:\\[2mm]
(a)\quad$\D\lim_{x\to 0^{+}}\frac{\sqrt{x}-x}{\sqrt{2x}-2x}$\qquad
(b)\quad$\D\lim_{x\to\infty}\frac{\sqrt{x}-x}{\sqrt{2x}-2x}$\qquad
(c)\quad$\D\lim_{x\to 0}\frac{x-\sin 3x}{x^{2}+\sin 2x}$
}
\LSG{
\begin{enumerate}
\item $\D\lim_{x\to 0^{+}}\frac{\sqrt{x}-x}{\sqrt{2x}-2x}=
  \lim_{x\to 0^{+}}\frac{1-\sqrt{x}}{\sqrt{2}-2\sqrt{x}}=
  \frac{1}{\sqrt{2}}$
\item $\D\lim_{x\to\infty}\frac{\sqrt{x}-x}{\sqrt{2x}-2x}=
  \lim_{x\to\infty}\frac{\frac{1}{\sqrt{x}}-1}
  {\frac{\sqrt{2}}{\sqrt{x}}-2}=\frac{1}{2}$
\item $\D\lim_{x\to 0}\frac{x-\sin 3x}{x^{2}+\sin 2x}=
  \left(\frac{0}{0}\right)=
  \lim_{x\to 0}\frac{1-3\cos 3x}{2x+2\cos 2x}=
  \frac{1-3}{2}=-1$
\end{enumerate}
}
