1
$\begingroup$

Suppose that $f$ is continuous with $\lim_{x\to\infty}f(x+1) - f(x) = 0$. Moreover $\lim_{n\to\infty} f(n)/n = 0$. How can I show that $\lim_{x\to\infty}f(x)/x = 0$ also?

(Only thing I can think of is trying to show that $f(x)$ is bounded, and then try to use the boundedness of $f(x)$ to come up with a proof. But then why would they bother prove $f(n)/n \to 0$ as $n \to \infty$? I feel like I am missing something obvious here.)

$\endgroup$
3
$\begingroup$

Let $c>0$, there exists $X>0$ such that $x>X$ implies that $|f(x+1)-f(x)|<c/2$, let $y>X$, there exists $x\in [X,X+1]$ and an integer $n$ such that $y=x+n$, $|f(y)-f(x)|\leq |f(x+n)-f(x+n-1)|+...+|f(x+1)-f(x)|\leq nc/2$, we deduce that $|f(y)|\leq |f(x)|+nc/2$ and $|f(y)/y|\leq |f(x)|/y+nc/2y\leq |f(x)|/n+c/2$, since $f$ is continuous, there exists $M$ such that for every $z\in [X,X+1], |f(z)|<M$, we deduce that $|f(y)|<M/n+c/2$, take $y>X$ such that $M/y<c/2$, we deduce that $|f(y)/y|<c$.

| cite | improve this answer | |
$\endgroup$

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.