Indeterminate form 0/0

Let's suppose that $lim_{x \to + \infty} f (x) = 0$ and $lim_{x \to + \infty} g (x) = 0$ , then we have that $lim_{x \to + \infty} \frac{f (x)}{g (x)} = \frac{0}{0}$ 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 $lim_{x \to + \infty} \frac{f (x)}{g (x)} = \frac{0}{0}$ implies that $lim_{x \to + \infty} \frac{f (x)}{g (x)} = lim_{x \to + \infty} \frac{\frac{1}{g (x)}}{\frac{1}{f (x)}} = \frac{\pm \infty}{\pm \infty}$ so we have changed the indeterminate form zero over zero by the one we already know, infinity over infinty.

Let's see some examples:

Example

a) $lim_{x \to + \infty} \frac{\frac{x}{x^{2} - 1}}{\frac{2 + x}{x^{2}}} = \frac{0}{0} \Rightarrow lim_{x \to + \infty} \frac{\frac{x}{x^{2} - 1}}{\frac{2 + x}{x^{2}}} = lim_{x \to + \infty} \frac{x \cdot x^{2}}{(x^{2} - 1) \cdot (2 + x)} = lim_{x \to + \infty} \frac{x^{3}}{x^{3}} = 1$

since $\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a \cdot b}{b \cdot c}$

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