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How can I compute $\lim_{x \to \infty} \frac{\log(x)}{x^a}$ for some $a \in \mathbb R$ with $x^a := e^{a \log(x)}$? Can you give me a hint?

I want to use only the basic properties of limits, like the linearity, multiplicativity, monotonicity, the Sandwich property and continuity (no L'Hospital, derivatives, integrals).

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  • $\begingroup$ it's 0 for $a>0$ and, obviously, is infinity for $a<0$. $\endgroup$ – Vasya Jan 5 '18 at 16:59
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For $a\leq 0$ the limit is clearly $+\infty$.

For $a> 0$, let $x=e^y \quad y\to +\infty$

$$\frac{\log x}{x^a}=\frac{\log e^y}{e^{ay}}=\frac{y}{e^{ay}}\leq\frac{y}{y^2}=\frac1y\to0$$

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Set $y:= a \cdot \log(x)$ than it follows for $a>0$ that

$$\lim_{x \rightarrow \infty} \frac{\log(x)}{x^{a}}= \lim_{x \rightarrow \infty} \frac{\log(x)}{e^{a \log(x)}} = \frac{1}{a} \lim_{y \rightarrow \infty} \frac{y}{e^{y}}=0$$

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