How is $\frac{\big(\frac{3}{2}\big)^{99}-1}{\big(\frac{3}{2}\big)^{100}-1}\approx\frac{1}{\big(\frac{3}{2}\big)}$ I read somewhere that $$\frac{\big(\frac{3}{2}\big)^{99}-1}{\big(\frac{3}{2}\big)^{100}-1}\approx\frac{1}{\big(\frac{3}{2}\big)}$$I don't  know how to have it. Please let me know how this is approximated.

Let $x>1$ and $q_n = \frac{x^n-1}{x^{n+1}-1}$.
We see that $q_n = \frac{1-\frac{1}{x^n}}{x-\frac{1}{x^n}}$, from which we have $\lim_n q_n = \frac{1}{x}$.
In this case we have $x= \frac{3}{2}$, hence $\lim_n \frac{(\frac{3}{2})^n-1}{(\frac{3}{2})^{n+1}-1} = \frac{1}{(\frac{3}{2})}$.
$n=99$ is sufficiently 'large' that the approximation works.
