0
$\begingroup$

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).

$\endgroup$
3
$\begingroup$

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 $\frac{g(x) - (g(x) - f(x))}{g(x)}$.

$\endgroup$
  • 1
    $\begingroup$ It can be even worse. $|f(x)-g(x)|$ may be unbounded: Try $f(x)=x+\sqrt x$, $g(x)=x$. $\endgroup$ – Ted Shifrin Oct 28 '13 at 20:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.