The logarithm is slower than any power

1. The inequality $\ln t\le t-1$ (the graph of $\ln$ lies under its tangent at $1$) holds for all $t>0$. For $x>1$,
$$\ln x=2\ln\sqrt x\le2\left(\sqrt x-1\right)<2\sqrt x.$$
Dividing by $x$,
$$0<\frac{\ln x}{x}<\frac{2}{\sqrt x}\longrightarrow0,$$
so $\dfrac{\ln x}x\to0$ by the squeeze theorem.

2. Let $a>0$ and set $y=x^a$. As $x\to+\infty$, $y\to+\infty$. Since $\ln x=\frac1a\ln y$,
$$\frac{\ln x}{x^a}=\frac1a\cdot\frac{\ln y}{y}\longrightarrow\frac1a\cdot0=0$$
by part 1.

Why the clever step? Applying $\ln t\le t-1$ directly to $t=x$ would only give $\ln x<x$, which is not enough. Using $t=\sqrt x$ gains a square root, and that is exactly what we need.