Let $f_n(x) : \mathbb{R} \rightarrow \mathbb{R}$ be defined as $\dfrac{x}{n}$. Is $f_n$ uniformly convergent?
With "$(a_n)$ is limited and $(b_n)$ converges to 0 $\Rightarrow (a_nb_n)$ converges also to 0" i would guess that $x\cdot \dfrac{1}{n}$ converges uniformly but $x$ is not limited in this case. So does ist converge uniformly or not?
If you are taking $x \in [a, b]$ a fixed, closed sub-interval of $\mathbb{R}$, then $|\frac{x}{n}| \leq |\frac{b}{n}|$, which can be made arbitrarily small only depending on the value of $n$.
If you are taking $x \in \mathbb{R}$, then given any $\epsilon > 0$, there exists $x \in \mathbb{R}$ such that $|\frac{x}{n}| \geq \epsilon$ (it is enough to choose $x \geq n \epsilon$). So you cannot have $|\frac{x}{n}| \leq \epsilon$ for all $x$ as soon as $n$ is big enough. Hence the convergence is not uniform on $\mathbb{R}$.
It really depends on which space you are considering your set of functions: on every bounded set the convergence will be uniform, in every unbounded, not uniform.