# Growth rate of 1 <mrow class="MJX-TeXAtom-ORD"> / </mrow> ( log &#x2061;<!--

Growth rate of $1/\left(\mathrm{log}\left(x\right)-\mathrm{log}\left(x-1\right)\right)$
Let $x>1$ be a real number. Let $y=\frac{1}{\mathrm{log}\left(x\right)-\mathrm{log}\left(x-1\right)}$
My question: Approximately how fast does $y$ grow (asymptotically) in terms of $x$? (e.g. linear, polynomial, exponential)?
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vrtuljakc6
$\mathrm{log}\left(b\right)-\mathrm{log}\left(a\right)$
$={\int }_{a}^{b}\frac{d\mathrm{log}\left(x\right)}{dx}dx$
$={\int }_{a}^{b}\frac{1}{x}dx$
Since $a, 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,
$\frac{b-a}{b}<\mathrm{log}\left(b\right)-\mathrm{log}\left(a\right)<\frac{b-a}{a}$
Putting $a=x-1$ and $b=x$ gives
$\frac{1}{x}<\mathrm{log}\left(x\right)-\mathrm{log}\left(x-1\right)<\frac{1}{x-1}$
So the answer is that $y$ is about $x$
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dikcijom2k
There is another way to look at this problem. If "x" is large, the function can writey = - 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.
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