Uniform continuity of $\frac{\sin (y^3)}{5+y}$ for $y \geq 0$ Is $f\colon y\mapsto\dfrac{\sin(y^3)}{5+y}$ for $y\geq 0$ uniformly continuous? I think it is but I can't seem to prove that. (Brain-dead day.) Help would be very much appreciated.
 A: I give the proof of the following result:

Let $f\colon \left[0,+\infty\right(\to \mathbb R$ a continuous function If $\displaystyle\lim_{x\to +\infty}f(x)=0$ then $f$ is uniformly continuous on $\left[0,+\infty\right($. 

Let $\varepsilon>0$. We can find $x_0\geq 0$ such that if $x\geq x_0$ then $|f(x)|\leq\frac{\varepsilon}2$. Since the interval $\left[0,x_0+1\right]$ is compact and $f$ is continuous, we can find a $0<\delta<1$ such that for all $x,y\in\left[0,x_0+1\right]$ which satisfy $|x-y|\leq \delta$ we have $|f(x)-f(y)|\leq\varepsilon$. Now, if we take $x,y\geq 0$ such that $|x-y|\leq\delta$, by symmetry, we have to look at three cases:
1) $x,y\in\left[0,x_0\right]$: we use the uniform continuity on $\left[0,x_0+1\right]$ to get that $|f(x)-f(y)|\leq\varepsilon$.
2) $x>x_0$, $y\leq x_0$: we should have $x\in \left[0,x_0+1\right]$ since $\delta<1$ and we use the uniform continuity on $\left[0,x_0+1\right]$.
3) $x>x_0$, $y>x_0$: we get $|f(x)-f(y)|\leq |f(x)|+|f(y)|\leq 2\frac{\varepsilon}2=\varepsilon$.
Now, we apply this to $x\mapsto \frac{\sin (x^3)}{x+5}$, and the following graph shows the behavior of this function. 

