Show $\frac{\sin(x^{3})}{x}$ is uniformly continuous on $[0,\infty)$. Show $f(x) = \frac{\sin(x^{3})}{x}$ is uniformly continuous on $[0,\infty)$.
I have the following two results to assist me:


*

*If $f$ is continuous on $(a,c)$ and u.c on $(a,b]$ and $[b,c)$, then it is u.c on $(a,c)$.

*If $\lim_{x \to 0^{+}}f(x)$ exists then $f(x)$ is u.c on $(0,1]$
Attempt:
Using the second hint and after applying L'Hopital's rule we will see that $\frac{\sin(x^{3})}{x} = 0$. As such I could turn this into a compact set: $[0,1]$ and uniform continutiy is given over such a space..
My Problem is in establishing u.c on $[1,\infty)$. Looking at hints to similar questions posted on here, I'm wondering if since we know the expression is continous over this interval then by the Mean Value Theorem there would exist a $\gamma > 0$ such that: $$|f(x) -f(y)| \leq |f'(\gamma)||x-y| < |f'(\gamma)| \delta$$
And then do the necessary manipulation on $\delta$. That is let $\delta = \frac{\epsilon}{|f'(\gamma)|}$ and then I would be done.
Concern: Is this the right procedure?
 A: You won't succeed in proving uniform continuity on an unbounded interval $[a, \infty)$ using MVT since the derivative
$$f'(x) = \frac{3x^2\cos x^3}{x}- \frac{\sin x^3}{x^2}  = 3x \cos x^3 - \frac{\sin x^3}{x^2},$$
is unbounded for large $x$.
However, we have $f(x) \to 0$ as $x \to \infty$.  For any $\epsilon > 0$ there exists $K > 0$ such that $|f(x)| < \epsilon/4 $ for all $x \geqslant K$. Thus, $|f(x) - f(y)| \leqslant |f(x)| + |f(y)| < \epsilon/2 < \epsilon$ for all $x, y \in [K, \infty)$. 
Note that $f$ can be extended continuously at $x=0$ by choosing $f(0) = 0$. Thus, $f$ is uniformly continuous on the compact interval $K$ and there exists $\delta(\epsilon)>0$ such that $|f(x) - f(y)| < \epsilon /2 $ for all $x,y \in [0,K]$ with $|x-y| < \delta(\epsilon)$.
Finally, if $x \in [0,K]$ and $y \in [K,\infty)$ with $|x-y| < \delta(\epsilon)$, we have 
$$|f(x) - f(y)| \leqslant |f(x) - f(K)| + |f(K)- f(y)|$$
The first term on the RHS is smaller than $\epsilon/2$ since $|x-K| \leqslant |x-y| < \delta(\epsilon)$.  The second term on the RHS is also smaller than $\epsilon/2$ since $K,y \in [K, \infty)$.
In conclusion we see that if $x,y \in [0,\infty)$ and $|x-y|< \delta(\epsilon)$ then we have $|f(x) - f(y)| < \epsilon$, proving uniform continuity. 
