I want to prove that $1=0.999...$ in this way, we have
$$0.9...9=0.9 \left(10^{-n-1}\right)\left(\sum_{i=0}^n 10^i\right)$$ Hence $$0.999...=\lim_{n \to \infty} 0.9 \left(10^{-n-1}\right)\left(\sum_{i=1}^n 10^i\right) \\ =0.9 \lim_{n \to \infty} \frac{\sum_{i=0}^n 10^i}{10^{n+1}}$$ Both the numerator and the denominator goes to $\infty$ so we xan use hôpital but even if we did we’d get $\infty/\infty$ again, how can I evaluate this limit?