$\lim_{x \to +\infty} f(x)g(x) = +\infty$. True or false?

Question

True or false? If true, justify. If false, give counter-example. If $f,g : \mathbb{R} \to \mathbb{R}$ are functions such that $f$ is bounded and positive and $\lim_{x \to +\infty} g(x) = +\infty$, so $\lim_{x \to +\infty} f(x)g(x) = +\infty$.

According to the answer key (I took a look at it because I had no idea how to answer this), that statement is false. So, let $f(x)=\sin{x}+1$. What about $g(x)$? No function came to my mind yet to serve as a counter-example. Any hint?

Answer 1

Counter example: $f(x)=e^{-x^2},g(x)=x^2.$

Answer 2

Let $g(x)$ be anything where where $\lim_{x\to 0} g(x) = \infty$ (for example $g(x) = x$). Let $f(x)$ be any function so that $f(x)$ is generally $\le \frac 1{g(x)}$. Okay we mus make some conditions so that $f$ is always bounded and positive, so for example $f(x) = \begin{cases}1 & g(x) < 1\\ \frac 1{g(x)}& g(x) \ge 1\end{cases}$.

The $f(x)$ is bounded ($ 0 < f(x) \le 1$)$ and so....

So $f(x)g(x)= \begin{cases} g(x) & g(x) < 1\\ 1 & g(x) \ge 1\end{cases}$

And $\lim_{x \to \infty}f(x)g(x) \ne \infty$.

In fact, I'll leave it to you to show that $\lim_{x\to \infty}f(x)g(x) = \lim_{x\to \infty}\frac 1{g(x)}g(x) = 1$. (Because there is an $M$ so that $g(x) > 1$ for all $x > M$).