Indeterminate form 0 x infinity
We will suppose that $\displaystyle\lim_{x \to{+}\infty}{f(x)}=0$ and $\displaystyle\lim_{x \to{+}\infty}{g(x)}= \pm \infty$, then we will have that $\displaystyle\lim_{x \to{+}\infty}{f(x) \cdot g(x)}= 0 \cdot \pm (\infty)$.
In other words, we are wondering what function goes more rapidly to its limit, $f(x)$ to zero or $g (x)$ to infinity.
To solve this type of indeterminate form we will do a simple step: $$\displaystyle\lim_{x \to{+}\infty}{f(x) \cdot g(x)}= \displaystyle\lim_{x \to{+}\infty}{\frac{1}{\frac{1}{f(x)}} \cdot g(x)}= \displaystyle\lim_{x \to{+}\infty}{\frac{g(x)}{\frac{1}{f(x)}}}=\frac{\pm \infty}{\pm \infty}$$ and we will solve the limit.
Let's see some examples:
- $\displaystyle\lim_{x \to{+}\infty}{\frac{2x}{x^3-1}\cdot \ln x}=0 \cdot (+ \infty) \Rightarrow \displaystyle\lim_{x \to{+}\infty}{\frac{2x}{x^3-1}} \cdot \ln x=\displaystyle\lim_{x \to{+}\infty}{\frac{2x \ln x}{x^3-1}}=$
$$=\displaystyle\lim_{x \to{+}\infty}{\frac{2x \cdot \ln x}{x^3}}=\displaystyle\lim_{x \to{+}\infty}{\frac{2 \cdot \ln x}{x^2}}=0$$
$\displaystyle\lim_{x \to{+}\infty}{x^{-x} \cdot 2^x}=\displaystyle\lim_{x \to{+}\infty}{\frac{2^x}{x^x}}=0$
$\displaystyle\lim_{x \to{+}\infty}{\frac{\ln x}{x +1} \cdot \frac{-x^2-1}{\ln x-1}}=\displaystyle\lim_{x \to{+}\infty}{\frac{\ln x \cdot (-x^2-1)}{(x+1) \cdot (\ln x-1)}}=\displaystyle\lim_{x \to{+}\infty}{-x}=- \infty$