How could I prove via limit definition that from $$ \lim_{n \to \infty} \frac{f(n)}{g(n)} = 1 $$ derives $$ \left| f(n) - g(n) \right| \to 0 $$ ?
Previous attempt took me to $$ \left| f(n) - g(n) \right| < \varepsilon \left| g(n) \right| \qquad \forall \varepsilon \space \text{fixed}, \forall n \ge n_0 $$ which is not useful since $\varepsilon \cdot |g|$ might be not that low ceil (...I'm thinking the case when $g$ goes to infinity).
Are you sure you have the right question? Consider the functions $f(x) = x+1$, and $g(x) = x$.
If you assume that $|f(x) - g(x)| \rightarrow 0$, it's not terribly hard to show that their ratio tends to 1.
Hint: rewrite $\frac{f(x)}{g(x)}$ as $\frac{g(x) - (g(x) - f(x))}{g(x)}$.