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Let $p>0$. If we know that $\displaystyle\lim_{x\to+\infty}\frac{x^{p}}{e^{x}}=0$, how can we show that $\displaystyle\lim_{x\to+\infty}\frac{\ln(x)}{x^{p}}=0$? My best guess is to use $x=e^{y}$ and $y=\ln(x)$ so I end up with $\displaystyle\lim_{y\to+\infty}\frac{y}{(e^{y})^{p}}$ which is not what I expected. I hope someone can help me out :)

Maria Kuznetsov
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  • Why? Aren't you done now? – Hagen von Eitzen Nov 22 '20 at 10:09
  • The problem is I don't know if I am done or not but I noticed that the limit I obtained is different from $\displaystyle\lim_{x\to+\infty}\frac{x^{p}}{e^{x}}=0$ which I want to use. That is to say, I wanted to obtain something like $\displaystyle\lim_{y\to+\infty}\frac{y^{p}}{e^{y}}=0$ or similar to it – Maria Kuznetsov Nov 22 '20 at 10:10
  • To get a limit which is exactly what you want to use, you can make the substitution $y = (\ln(x))^{1/p}$. Or you could make a second substitution $y = z/p$. However, it's important to understand that $y/e^{yp} \to 0$ when $p>0$ for the same reason that $x^p/e^x$ does - if that isn't obvious to you, I recommend trying to prove both of them and looking at the similarities. – preferred_anon Nov 22 '20 at 10:39

3 Answers3

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Let $p\gt0$. Using L'Hopital's Rule:

$$\lim_{x\to+\infty}\frac{\ln(x)}{x^{p}} = \lim_{x\to+\infty} \frac{1/x}{px^{p-1}} = \lim_{x\to+\infty} \frac{1}{px^p} = 0$$

For other methods check this.

S.H.W
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Can you use continuity? if yes then: $$\frac{y}{(e^{y})^{p}}=\left(\frac{y^{\frac{1}{p}}}{e^{y}}\right)^p$$

zkutch
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$p>0;$

$(1/p)\dfrac{\log (x^p)}{x^p};$

Set $y:=x^p$ and consider $y \rightarrow \infty.$

($1/p)\lim_{y \rightarrow \infty}\dfrac{\log y}{y}=?$

Peter Szilas
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