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

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 $a<0$. – Vasya Jan 5 '18 at 16:59

For $a\leq 0$ the limit is clearly $+\infty$.
For $a> 0$, let $x=e^y \quad y\to +\infty$
$$\frac{\log x}{x^a}=\frac{\log e^y}{e^{ay}}=\frac{y}{e^{ay}}\leq\frac{y}{y^2}=\frac1y\to0$$
Set $y:= a \cdot \log(x)$ than it follows for $a>0$ that
$$\lim_{x \rightarrow \infty} \frac{\log(x)}{x^{a}}= \lim_{x \rightarrow \infty} \frac{\log(x)}{e^{a \log(x)}} = \frac{1}{a} \lim_{y \rightarrow \infty} \frac{y}{e^{y}}=0$$