# Spivak Chapter 22 Problem 17

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.)

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 $|f(y)/y|<c$.