$$f(x) = \frac{\sin x^3}{x+1}$$
If $f(x)$ is defined for $x \in [0,\infty)$, I can see that its derivative is bounded in the interval so it is just the matter of proving it. Im going for an $e-s$ proof, but I'm stumbling in terms of finding an appropriate $s.$ Also, would be nice if someone could give an explanation if possible what it means for a function to be uniformly continuous graphically - such that I can tell almost instantly if a function is uniform continuous (if possible).
Thanks.
Take $\epsilon > 0$. Find $k > 1$ large enough that $1/x < \epsilon/4$ whenever $x \geq k$. Now, since $f(x)$ is continuous and $[0,k+1]$ is compact, then $f$ restricted to $[0,k+1]$ is uniformly continuous. Let $\delta_1 > 0$ be small enough that if $x, y \in [0,k+1]$ then $|x-y|< \delta_1$ implies $|f(x)-f(y)|<\epsilon$. Now, suppose that $x \in (k+1,\infty)$ and $|x-y|<1$. In particular this implies that $k<x$ and $k<y$. Then $$ |f(x)-f(y)| = \left| \frac{(y+1) \sin(x^3) - (x+1)\sin(y^3)}{(x+1)(y+1)}\right| \\ \leq \left| \frac{y\sin(x+1) - x\sin(y+1)}{(x+1)(y+1)}\right| + \left|\frac{\sin(x+1)-\sin(y+1)}{(x+1)(y+1)} \right| \\ \leq \frac{2 \max(x,y)}{(x+1)(y+1)}+\frac{2}{(x+1)(y+1)} \\ < \frac{2}{\min(x,y)}+\frac{2}{(x+1)(y+1)} \\ < 2\frac{1}{k}+2\frac{1}{k} \\ < 4(\epsilon/4) = \epsilon. $$ The very last inequality came with our choice of $k$. So, taking $\delta = \min(\delta_1, 1)$ will do the trick!
Graphically, you might think of uniform continuity as saying that $f$ can't move vertically too steeply at any points (otherwise any small changes in the domain [controlled by $\delta$] can still result in large changes in the range [controlled by $\epsilon$]). For example, $f(x) = 1/x$ is not uniformly continuous on $(0,1)$ since near $0$ it grows nearly vertically. Similarly, for any $n > 1$, $f(x) = \pm x^n$ is not uniformly continuous on $(0, \infty)$ because as $x$ approaches $\infty$ the function $f$ moves nearly vertically (up if $+x^n$ down if $-x^n$).