> Ich will also limh->0 ((x0+h)-x0)/h herausbekommen.
Ich nehme an, du meinst limh→0 \(\frac{f(x_0+h)-f(x_0)}{h}\) = limh->0 \(\frac{1/(x_0+h)-1/x_0}{h}\) (#)
Es gilt:
\(\frac{1/(x_0+h)-1/x_0}{h}\) = \(\frac{1}{h}\) • (\(\frac{1}{x_0+h}\)+ \(\frac{1}{x_0}\)) = \(\frac{1}{h}\) • \(\frac{x_0-(x_0+h)}{x_0·(x_0+h)}\) = \(\frac{1}{h}\) • \(\frac{-h}{x_0·(x_0+h)}\) = \(\frac{-1}{x_0·(x_0+h)}\)
→ (#) = limh→0 \(\frac{-1}{x_0·(x_0+h)}\) = \(\frac{-1}{x_0^2}\)
Gruß Wolfgang