# Proving that if $f$ is equal to $g$ asymtotically then their distance tends to zero

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 $\frac{g(x) - (g(x) - f(x))}{g(x)}$.
• It can be even worse. $|f(x)-g(x)|$ may be unbounded: Try $f(x)=x+\sqrt x$, $g(x)=x$. – Ted Shifrin Oct 28 '13 at 20:40