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