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.

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?

Are you sure you don't mean $\lim_{x\to\infty}$? The $\lim_{x\to0}$ doesn't even exist (although $\lim_{x\to0^+}$ doesn't), but e.g. $\lim_{x\to0^+}\frac{\ln x}{x}=-\infty$. — J.G., Jan 31 '22 at 15:43
First guess is $\ln (a) = - \ln(1/a)$. Then I think the proof might otherwise still apply. — TurlocTheRed, Jan 31 '22 at 15:43
Use $\frac{\ln x}{x^a}=\frac{2\ln x^{a/2}}{ax^a}\le\frac{2}{ax^{a/2}}$. — J.G., Jan 31 '22 at 15:48

Peter Szilas · Accepted Answer · 2022-02-02T03:48:37.953

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.

edited Feb 02 '22 at 03:48

answered Jan 31 '22 at 16:17

Peter Szilas

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

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