# How to prove $\lim_{x \to \infty}\frac{\log(x)}{x}=0$? [duplicate]

I will soon make a math exame where one can't use L'Hôpital's rule, integrals concept neither the formal limit definition. The most I can use is the derivative definition and the algebraic ways to solve limits.

My thought was:

Let $\displaystyle \lim_{x \to \infty} \frac{\log(x)}{x}=y$, so $\displaystyle \lim_{x \to \infty}\log(x^{\frac{1}{x}})=y$.

Then,

$$\displaystyle \lim_{x \to \infty}e^{\log(x^{\frac{1}{x}})}=e^y \Leftrightarrow$$

$$\displaystyle \lim_{x \to \infty}x^{\frac{1}{x}}=e^y$$

I ended with an indetermination.

How can I proof the $\displaystyle \lim_{x \to \infty}\frac{\log(x)}{x}=0$ with the restritions and as formal it can get? Thanks.

## marked as duplicate by Did limits StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 23 '15 at 17:10

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• Then you need to say what you can use, for example how you define the logarithm. – Did May 23 '15 at 17:08
• Look back maybe?(it was you who asked about it) – JukesOnYou May 23 '15 at 17:10
• @Did The question specified the prohibition of L'Hospital's Rule. The answers to the "already asked question" used L'Hospital's Rule. – Mark Viola May 23 '15 at 17:31

## 1 Answer

Write $\log(x)=y$ then $x=e^y$ and as $x\to\infty$, $y\to\infty$.

You want to find $$\lim_{y\to\infty}\dfrac{y}{e^y}$$

But $e^y\geq 1+y+y^2/2!$ for any $y\in\mathbb R$ hence $$0\leq\lim_{y\to\infty}\dfrac{y}{e^y}\leq\lim_{y\to\infty}\dfrac{y}{1+y+y^2/2!}=0$$ so $$\lim_{y\to\infty}\dfrac{y}{e^y}=0$$