Let's suppose that $$\displaystyle\lim_{x \to{+}\infty}{f(x)}=0$$ and $$\displaystyle\lim_{x \to{+}\infty}{g(x)}=0$$, then we have that $$\displaystyle\lim_{x \to{+}\infty}{\frac{f(x)}{g(x)}}=\frac{0}{0}$$ and thus there is an indeterminate form.
In this case we have to ask ourselves what function tends more rapidly toward zero: if it is $$f(x)$$, then the limit will be zero, and if it is $$g(x)$$ then the limit will be infinity.
Note that $$\displaystyle\lim_{x \to{+}\infty}{\frac{f(x)}{g(x)}}=\frac{0}{0}$$ implies that $$\displaystyle\lim_{x \to{+}\infty}{\frac{f(x)}{g(x)}}=\displaystyle\lim_{x \to{+}\infty}{\frac{\frac{1}{g(x)}}{\frac{1}{f(x)}}}=\frac{\pm \infty}{\pm \infty}$$ so we have changed the indeterminate form zero over zero by the one we already know, infinity over infinty.
Let's see some examples:
a) $$\displaystyle\lim_{x \to{+}\infty}{\frac{\frac{x}{x^2-1}}{\frac{2+x}{x^2}}}=\frac{0}{0} \Rightarrow \displaystyle\lim_{x \to{+}\infty}{\frac{\frac{x}{x^2-1}}{\frac{2+x}{x^2}}}=\displaystyle\lim_{x \to{+}\infty}{\frac{x \cdot x^2}{(x^2-1) \cdot (2+x)}}=\displaystyle\lim_{x \to{+}\infty}{\frac{x^3}{x^3}}=1$$
since $$\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a \cdot b}{b \cdot c}$$
b) $$\displaystyle\lim_{x \to{+}\infty}{\frac{\frac{2+x}{\ln x}}{\frac{\ln x}{x}}}=\displaystyle\lim_{x \to{+}\infty}{\frac{x \cdot (2+x)}{\ln x^2}}=\displaystyle\lim_{x \to{+}\infty}{\frac{x^2}{\ln x^2}}=+ \infty$$