Finding limit without using L'Hopital rule

Question

How to prove $$\lim_{x \to \infty} \frac {\ln(x)}{x}=0$$ without using L'Hospital Rule. Just by using some basis limit properties.

Answer 1

For sufficiently large $x$:$0\le \ln (x)\le \sqrt x$. Now divide by $x$ and squeeze.

Answer 2

set $$x=e^t ,t\rightarrow \infty $$so now $$\lim_{x\rightarrow \infty }\frac {ln(x)}{x}=\\\lim_{t\rightarrow +\infty }\frac {ln(e^t)}{e^t}=\\\lim_{t\rightarrow +\infty }\frac {t}{e^t}=\\\\\lim_{t\rightarrow +\infty }\frac {t}{1+t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+....}=\\$$ obviously the limit go to zero

Answer 3

The Taylor series for $\ln(x)$ in $-1\le x \le 1$ is given by :

You can substitute this and then apply the basic limits results to get the answer.