Indeterminate form 0/0

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) $\displaystyle\lim_{x \to{+}\infty}{\frac{\frac{2+x}{\ln x}}{\frac{\ln x}{x}}}=\displaystyle\lim_{x \to{+}\infty}{\frac{x \cdot (2+x)}{\ln x^2}}=\displaystyle\lim_{x \to{+}\infty}{\frac{x^2}{\ln x^2}}=+ \infty$