How is $\frac{\big(\frac{3}{2}\big)^{99}-1}{\big(\frac{3}{2}\big)^{100}-1}\approx\frac{1}{\big(\frac{3}{2}\big)}$

Question

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.

Answer 1

Another point of view: $$(3/2)\left((3/2)^{99}-1\right)=(3/2)^{100}-(3/2)=(3/2)^{100}-1-0.5$$

Answer 2

$(3/2)^{99}$ and $(3/2)^{100}$ are both really big compared to the $-1$ term in the numerator and the denominator. So the idea is that you can ignore the $-1$ terms and evaluate the fraction as $(3/2)^{99} / (3/2)^{100} = 1/(3/2)$.

Answer 3

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.