the limit of ln(x) is infinite, so why does dividing it by x make it zero? If anything it should make it 1 because infinite/infinite = 1
$\displaystyle \lim_{x \to \infty}\frac{\ln{x}}{x}$.
Since this goes to $\displaystyle \frac{\infty}{\infty}$ you can use L'Hospital's Rule.
$\displaystyle \lim_{x \to \infty}\frac{\ln{x}}{x} = \lim_{x \to \infty}\frac{\frac{d}{dx}(\ln{x})}{\frac{d}{dx}(x) }$
$\displaystyle = \lim_{x \to \infty}\frac{\frac{1}{x}}{1}$
$\displaystyle = \lim_{x \to \infty}\frac{1}{x}$
$\displaystyle = 0$.
Even though both numerator and denominator are growing without bound, the real question is which is growing faster.
Also note that for $\displaystyle x \ge e$
$\displaystyle 1 \le \ln x \le \sqrt{x}$
so
$\displaystyle \dfrac{1}{x} \le \dfrac{\ln x}{x} \le \dfrac{1}{\sqrt{x}}$,
Thus,
$\displaystyle \displaystyle \lim_{x \to \infty} \dfrac{1}{x} \le \lim_{x \to \infty} \dfrac{\ln x}{x} \le \lim_{x \to \infty} \dfrac{1}{\sqrt{x}}$
which gives
$\displaystyle \displaystyle \lim_{x \to \infty} \dfrac{\ln x}{x}} = 0.$
A non L'Hopital way of doing it.