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I need to find a function $a(t)$ which is positive such that for some $\gamma > 0$ we have:

$$\lim_{{t \to \infty}}\frac1{a(t)}\left({-\frac{1}{{\ln(1-\frac{1}{{xt}})}} + \frac{1}{{\ln(1-\frac{1}{t})}}}\right) = \frac{x^{\gamma}-1}{\gamma} $$

I know by using Wolfram Alpha that $a(t) = t$ and $\gamma = 1$ suffice, i.e.

$$\lim_{{t \to \infty}} \frac1t\left({-\frac{1}{{\ln(1-\frac{1}{{xt}})}} + \frac{1}{{\ln(1-\frac{1}{t})}}}\right)= x-1 $$

But why is this true? Even knowing the correct function $a(t)$ I still don't see why this holds. I tried l'Hôpital and substitution but that only makes matters worse.

Anne Bauval
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user979120
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    Simplify denominator and use logarithm properties? – zkutch Jun 20 '23 at 19:53
  • @zkutch Do you mean the denominators of the fractures with the Logarithms? I know you can write $\ln(1-\frac{1}{xt}) = \ln(\frac{xt-1}{xt}) = \ln(xt-1) - \ln(xt)$ but I don't see how that helps. – user979120 Jun 20 '23 at 19:58

2 Answers2

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$$\frac{{-\frac{1}{{\ln(1-\frac{1}{{xt}})}} + \frac{1}{{\ln(1-\frac{1}{t})}}}}{{t}} = \frac{\ln(1-\frac{1}{t})-\ln(1-\frac{1}{{xt}})}{t\ln(1-\frac{1}{{xt}})\ln(1-\frac{1}{t})}\sim\frac{\ln(1-\frac{1}{t})-\ln(1-\frac{1}{{xt}})}{t\cdot\frac{1}{t}\cdot\frac{1}{{xt}}}\quad(1)$$

  1. Can you finish using $\ln(1+x)=x-\frac{x^2}{2}+o(x^2)$ for denominator?

  2. We can use more simple variant $\ln(1+x)\sim x, x\to 0$ and use $$\ln\left(1-\frac{1}{t}\right)-\ln\left(1-\frac{1}{{xt}}\right) = \ln\frac{x(t-1)}{xt-1}=\ln\left(1+\frac{1-x}{xt-1}\right)\sim \frac{1-x}{xt-1}$$

  3. l'Hôpital in $(1)$ also gives simple answer.

p.s. There is one little slight difference between my answer and your question..

zkutch
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Appply $$\lim_{u\to0}\frac{\ln(1+u)}u=\ln'(1)=1$$ to $u=-\frac1{xt}:$ $$\lim_{t\to\infty}-t\ln\left(1-\frac1{xt}\right)=\frac1x\tag A$$ In particular, $$\lim_{t\to\infty}-t\ln\left(1-\frac1t\right)=1\tag B$$ $\frac1{(A)}-\frac1{(B)}$ gives the expected result.

Anne Bauval
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