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


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

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

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 May 23 '15 at 17:10

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  • $\begingroup$ Then you need to say what you can use, for example how you define the logarithm. $\endgroup$ – Did May 23 '15 at 17:08
  • 1
    $\begingroup$ Look back maybe?(it was you who asked about it) $\endgroup$ – JukesOnYou May 23 '15 at 17:10
  • $\begingroup$ @Did The question specified the prohibition of L'Hospital's Rule. The answers to the "already asked question" used L'Hospital's Rule. $\endgroup$ – Mark Viola May 23 '15 at 17:31

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 $$\lim_{y\to\infty}\dfrac{y}{e^y}=0$$


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