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I don't know how to prove that $\lim_{x\to\infty}\frac{\ln x}{x-1}=0$ without using L'Hopital.

I've tried to use the definition of $\ln x=\int_{1}^{x}\frac{1}{t}dt$ and the fact that $\frac{x-1}{x}<\ln x<x-1$ but I don't get anything. Another thing I've tried is to prove the limit by definition but it's quite complicated. Any suggestions?

Iesus Dave Sanz
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  • You can maybe prove that for sufficiently large $x$ we have $\ln x < \sqrt{x}$? – Slugger Oct 22 '20 at 01:15
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    This isn't rigorous, but an intuition would be that polynomials grow faster than logarithmic functions, so the value of a logarithmic function compared to that of a polynomial becomes negligible for a sufficiently large $x$. – Sage Stark Oct 22 '20 at 01:17
  • @Slugger In fact, $\log(x)<\sqrt{x}$ for all $x>0$. – Mark Viola Oct 22 '20 at 03:05

2 Answers2

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We have $$e^{\sqrt{x}} > 1 + \sqrt{x} + \frac{x}{ 2} + \frac{x^{3/2}}{6} ,$$ so that for $x > 6^2$ we have $e^{\sqrt{x}} > x = e^{\ln x}$. Thus $\sqrt{x} > \ln x$ for $x>36$ and thus also $$ \frac{\ln x}{x-1} < \frac{\sqrt{x}}{x-1} $$ for large enough $x$. The limit of the right side is zero, and the limit of the left side is positive. So we are done

Slugger
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The integral formula $\ln x = \displaystyle \int_1^x \dfrac 1t \, dt$ gives you $0 \le \ln x \le x$ for all $x \ge 1$. Thus if $t \ge 1$ and $x = \sqrt t$ then $$0 \le \frac{\ln t}{t} = \frac{\ln x^2}{x^2} = \frac{2 \ln x}{x^2} \le \frac 2x = \frac 2{\sqrt t}.$$ From here the limit of $\dfrac {\ln t}{t-1}$ follows quickly.

Umberto P.
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