\ANG{
Ermitteln Sie folgende Grenzwerte.\\[2mm]
\begin{tabular}{lll}
(a) $\D\lim_{x \to \infty} \left( x^2\cdot \sin \frac{\D 1}{\D x} 
    \right)$ &
(b) $\D\lim_{x \to 0} \left( \frac{\D 1}{\D x}-
     \frac{\D 1}{\D \sin x}\right)$ &
(c) $\D\lim_{x \to \frac{\pi}{2}} 
     \left[ (x - \frac{\pi}{2})\tan x \right]$ 
\end{tabular}

}
\LSG{
\begin{enumerate}
\item $\D\lim_{x\to\infty}\frac{\D\sin\frac{1}{x}}{\D\frac{1}{x^{2}}}=
  \lim_{x\to\infty}\frac{\D-\frac{1}{x^{2}}\,\cos\frac{1}{x}}
  {\D-\frac{2}{x^{3}}}=
  \lim_{x\to\infty}\left(\frac{x}{2}\,\cos\frac{1}{x}\right)=+\infty$\\
\item $\D\lim_{x\to 0}\frac{\sin x-x}{x\,\sin x}=
  \lim_{x\to 0}\frac{\cos x-1}{\sin x+x\,\cos x}=
  \lim_{x\to 0}\frac{-\sin x}{2\,\cos x-x\,\sin x}=0$\\
\item $\D\lim_{x\to\frac{\pi}{2}}
  \frac{(x-\frac{\pi}{2})\cdot\sin x}{\cos x}=
  \lim_{x\to\frac{\pi}{2}}
  \frac{\sin x+(x-\frac{\pi}{2})\cdot\cos x}{-\sin x}=-1$  
\end{enumerate}
}
