$\lim_{x\to+\infty}\frac{\ln(x)}{x^{p}}$

Question

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 :)

Answer 1

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

Answer 2

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.

Answer 3

$p>0;$

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

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

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