Example 5. Find $\lim _{x \rightarrow \infty} \frac{\ln x}{x^{\alpha}}$ for $\alpha>0, x>0$.
Solution. Applying the L'Hôpital-Bernoulli rule, we get

$$
\lim _{x \rightarrow \infty} \frac{\ln x}{x^{\alpha}}=\lim _{x \rightarrow \infty} \frac{1}{x \alpha x^{\alpha-1}}=\lim _{x \rightarrow \infty} \frac{1}{\alpha x^{\alpha}}=0
$$

Thus,

$$
\lim _{x \rightarrow \infty} \frac{\ln x}{x^{\alpha}}=0 \text { for } \alpha>0
$$

or

$$
\lim _{x \rightarrow \infty} \frac{x^{\alpha}}{\ln x}=\infty
$$

Remark. The logarithmic function $\ln x$ grows more slowly than any positive power of $x$.