Growth rate of $1/(\log(x)-\log(x-1))$

Question

Let $x>1$ be a real number. Let $y=\dfrac{1}{\log(x)-\log(x-1)}$.

My question: Approximately how fast does $y$ grow (asymptotically) in terms of $x$? (e.g. linear, polynomial, exponential)?

Answer 1

$$\log(b) - \log(a)$$ $$= \int_a^b \frac{d\log(x)}{dx} dx$$ $$= \int_a^b \frac{1}{x} dx$$

Since $a<x<b$, we have $\frac{1}{b} < \frac{1}{x} < \frac{1}{a}$, so

$$\int_a^b \frac{1}{b} dx < \int_a^b \frac{1}{x} dx < \int_a^b \frac{1}{a} dx$$

Therefore,

$$\frac{b-a}{b} < \log(b)-\log(a) < \frac{b-a}{a}$$

Putting $a = x-1$ and $b = x$ gives

$$\frac{1}{x} < \log(x) - \log(x-1) < \frac{1}{x-1}$$

So the answer is that $y$ is about $x$.

Answer 2

There is another way to look at this problem. If "x" is large, the function can write
y = - 1 / Log[1 - 1 / x]. Computing a Taylor series around an infinite value of "x" leads to y = x - 1/2 - 1/ (12 x) - 1 / (24 x^2) + .... So, "y" varies linearly just as "x". You could easily check that this is a very good approximation for x=2. If you compute the second derivayive of "y" with respect to "x", you will also notice that its value is very small, then very low curvature.