Prove $ \frac{1}{x-1} \approx \frac{1}{x}, x \gg 1 $

Question

I often found this form (and another similar type). I can understand it intuitively, but I don't know how to prove it analytically

Answer 1

You can use $\dfrac {\frac1x}{\frac1{x-1}}=1-\frac1x \to 1$ as $x$ increases

Answer 2

What was probably intended is a Taylor expansion using the geometric series, assuming $|x|>1$, as follows. We have \begin{align} \frac{1}{x-1}&=\frac{1}{x}\cdot\frac{1}{1-\frac{1}{x}}\\ &=\frac{1}{x}\cdot\sum_{n=0}^{\infty}\left(\frac{1}{x}\right)^n \end{align} The last equal sign is justified since our hypothesis $|x|>1$ implies $\left|\frac{1}{x}\right|<1$, and thus the geometric series can be used. So, one has \begin{align} \frac{1}{x-1}&=\frac{1}{x}+\mathcal{O}\left(\left(\frac{1}{x}\right)^2\right) \end{align} i.e to first order in the quantity $1/x$, we have $\frac{1}{x-1}\approx \frac{1}{x}$. I should emphasize that there are several ways to interpret the vague (since we have no context) symbol $\approx$. In this answer, I have given you one possible way to interpret the approximation.