Originally Posted by

**emakarov** I don't agree with these two inequalities, but it is true that

$$\left|\frac{\sin(x^3)}{x}-\frac{\sin(y^3)}{y}\right|\le\frac{1}{x}+\frac{1}{ y}\le \frac{2}{\min(x,y)}.$$Then we can find an $x_0$ such that $2/x<\epsilon$ for all $x> x_0$. Thus, if $x,y>x_0$, then $|f(x)-f(y)|<\epsilon$ regardless of how far $x$ is from $y$. Concerning the closed segment $[1,x_0]$, the function is continuous and therefore uniformly continuous there, so for the given $\epsilon$ there exists a $\delta$ such that $|x-y|<\delta$ implies $|f(x)-f(y)|<\epsilon$. Therefore, this $\delta$ will work on the whole interval $[1,\infty)$. (To be strict, one also has to take care of the case $x<x_0<y$.)