\ANG{
Berechne folgende Grenzwerte:\\[2mm]
\begin{tabular}[t]{lll}
(a)\quad$\D\lim_{x\to 0}\frac{\sin ax}{\sin bx}$\qquad & 
(b)\quad$\D\lim_{x\to a}\frac{\sqrt{x}-\sqrt{a}}{x-a}$\qquad &
(c)\quad$\D\lim_{x\to a^{+}}\frac{\sqrt{x}-\sqrt{a}}{\sqrt{x-a}}$\rule[-4mm]{0mm}{12mm}\\
(d)\quad$\D\lim_{x\to +\infty}\frac{\sqrt{x}}{\ln x}$\qquad &
(e)\quad$\D\lim_{x\to +\infty}x^{100}\e^{-0{,}01x}$\qquad &
(f)\quad$\D\lim_{x\to +\infty}\left(\e^{-x}\cdot\ln x\right)$\rule[-4mm]{0mm}{12mm}\\
(g)\quad$\D\lim_{x\to +\infty}\left(x-\ln x\right)$\qquad &
(h)\quad$\D\lim_{x\to +\infty}\left(\frac{\e^{2x}}{x^{2}}-\frac{\e^{x}}{x}\right)$\qquad &
(i)\quad$\D\lim_{x\to +\infty}\frac{2x+\sin x}{2x-\sin x}$
\end{tabular}
} 
\LSG{
\begin{enumerate}
\item $\D\lim_{x\to 0}\frac{\sin ax}{\sin bx}=\left(\frac{0}{0}\right)=
\lim_{x\to 0}\frac{a\cos ax}{b\cos bx}=\frac ab$
\item $\D\lim_{x\to a}\frac{\sqrt{x}-\sqrt{a}}{x-a}=\left(\frac{0}{0}\right)=
\lim_{x\to a}\frac{\sqrt{x}-\sqrt{a}}{(\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a})}=
\lim_{x\to a}\frac{1}{\sqrt{x}+\sqrt{a}}=\frac{1}{2\sqrt a}$
\item$\D\lim_{x\to a}\frac{\sqrt{x}-\sqrt{a}}{\sqrt{x-a}}=\left(\frac{0}{0}\right)=
\lim_{x\to a}\frac{\sqrt{x}-\sqrt{a}}{\sqrt{\sqrt{x}-\sqrt{a}}\sqrt{\sqrt{x}+\sqrt{a}}}=
\lim_{x\to a}\sqrt{\frac{\sqrt{x}-\sqrt{a}}{\sqrt{x}+\sqrt{a}}}=0$
\item $\D\lim_{x\to +\infty}\frac{\sqrt{x}}{\ln x}=\left(\frac{\infty}{\infty}\right)=
\lim_{x\to +\infty}\frac{\frac 1{2\sqrt{x}}}{\frac 1x}=\lim_{x\to +\infty}\frac{\sqrt x}{2}=\infty$
\item $\D
\lim_{x\to +\infty}\frac{x^{100}}{\e^{0{,}01x}}=\left(\frac{\infty}{\infty}\right)
\stackrel{\text{100 mal de l'Hospital}}{=}
\lim_{x\to +\infty}\frac{100!}{0{,}01^{100}\e^{0{,}01x}}=0$
\item $\D\lim_{x\to +\infty}\frac{\ln x}{\e^{x}}=\left(\frac{\infty}{\infty}\right)=
\lim_{x\to +\infty}\frac{\frac 1x}{\e^{x}}=\lim_{x\to +\infty}\frac{1}{x\e^{x}}=0$
\item $\D\lim_{x\to +\infty}\left(x-\ln x\right)=
\lim_{x\to +\infty}x\left(1-\frac{\ln x}x\right)=
\lim_{x\to +\infty}x\left(1-\frac{1}x\right)=\infty$
\item $\D\lim_{x\to +\infty}\left(\frac{\e^{2x}}{x^{2}}-\frac{\e^{x}}{x}\right)=
\lim_{x\to +\infty}\frac{\e^{x}}{x}\left(\frac{\e^{x}}{x}-1\right)=(\infty\cdot\infty)=\infty$
\item $\D\lim_{x\to +\infty}\frac{2x+\sin x}{2x-\sin x}=
\lim_{x\to +\infty}\frac{2+\frac{\sin x}{x}}{2-\frac{\sin x}{x}}=\frac 22=1$

\end{enumerate}
}
