# Limit at infinity of $\frac{\ln x}{x-1}$

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 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?

• You can maybe prove that for sufficiently large $x$ we have $\ln x < \sqrt{x}$? Oct 22, 2020 at 1:15
• 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 $x$. Oct 22, 2020 at 1:17
• @Slugger In fact, $\log(x)<\sqrt{x}$ for all $x>0$. Oct 22, 2020 at 3:05

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
• In fact, $\log(x)<\sqrt{x}$ for all $x>0$. Oct 22, 2020 at 3:02
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.
• In fact, $\log(x)<\sqrt {x}$ for all $x>0$. Oct 22, 2020 at 3:04