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

Przemysław Scherwentke
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aavatar
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3 Answers3

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

Ittay Weiss
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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

Khosrotash
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  • i got lim t -> infinity t/(e^t). What now?? please... – aavatar Nov 14 '14 at 08:55
  • estimate $e^t$ by some of the first terms of the taylor series. You should be able to obtain an estimate of the form $0\leq t/e^t\leq c/t^\alpha$ for some $\alpha\geq 1$, $c>0$ and $t$ large enough. – Quickbeam2k1 Nov 14 '14 at 09:17
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The Taylor series for $\ln(x)$ in $-1\le x \le 1$ is given by :

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

creative
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