Compute $\lim_{x \to \infty} \frac{\log(x)}{x^a}$

Question

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).

it's 0 for $a>0$ and, obviously, is infinity for $a<0$. – Vasili Jan 05 '18 at 16:59 — Vasili, Jan 05 '18 at 16:59

score 2 · Accepted Answer · answered Jan 05 '18 at 16:58

2

For $a\leq 0$ the limit is clearly $+\infty$.

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

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

answered Jan 05 '18 at 16:58

user

154,566

score 1 · Answer 2 · answered Jan 05 '18 at 17:02

1

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

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

answered Jan 05 '18 at 17:02

McBotto.t

452

Compute $\lim_{x \to \infty} \frac{\log(x)}{x^a}$

2 Answers2

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