Prove that $\lim_{x \to \infty} \frac{\ln(x)}{x^a} = 0$ for $a>0$

Question

Prove that $$\lim_{x \to \infty} \frac{\ln(x)}{x^a} = 0$$ for $a>0$ without using L'Hôpital.

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With the estimate $\ln(x) \leq x$ and the squeeze theorem I was able to show that if $a > 1$ this is indeed the case. I couldn't come up with a proof for $0 < a < 1$. Any suggestions?

Answer 1

Set $y:=x^a$, and consider $\lim y \rightarrow \infty$.

We then have $(1/a)(\log y) /y.$

$0<(\log y) /y=(1/y)\displaystyle{\int_{1}^{y}} (1/t)dt <$

$(1/y)\displaystyle{\int_{1}^{y}} 1/(\sqrt{t})dt=(1/y)2t^{1/2} \bigg ]_1^y=$

$(2/y)(y^{1/2}-1)<(2/y)(y)^{1/2}$

Take the limit.