How to calculate $ \lim_{s \to \infty} \frac{ab + (ab)^2 + ... (ab)^s}{1 +ab + (ab)^2 + ... (ab)^s} $ I'm trying to calculate this limit expression:
$$ \lim_{s \to \infty} \frac{ab + (ab)^2 + ... (ab)^s}{1 +ab + (ab)^2 + ... (ab)^s} $$
Both the numerator and denominator should converge, since $0 \leq a, b \leq 1$, but I don't know if that helps. My guess would be to use L'Hopital's rule and take the derivative with respect to $s$, which gives me:
$$ \lim_{s \to \infty} \frac{s (ab)^{s-1}}{s (ab)^{s-1}} $$
but this still gives me the non-expression $\frac{\infty}{\infty}$ as the solution, and applying L'Hopital's rule repeatedly doesn't change that. My second guess would be to divide by some multiple of $ab$ and therefore simplify the expression, but I'm not sure how that would help, if at all. 
Furthermore, the solution in the tutorial I'm working through is listed as $ab$, but if I evaluate the expression that results from L'Hopital's rule, I get $1$ (obviously). 

The derivative (with respect to $x$) of $a^x$ is not $x a^{x-1}$ but $ln(a)a^x$, since $a^x=e^{x ln(a)}$.
You can solve your problem by noticing that $\displaystyle \frac{ab + (ab)^2 + ... (ab)^s}{1 +ab + (ab)^2 + ... (ab)^s}=1- \frac{1}{1 +ab + (ab)^2 + ... (ab)^s}$.
