Proof that the function is uniformly continuous 
Proof that the function $f: [0, \infty) \ni  x \mapsto \frac{x^{2}}{x + 1} \in \mathbb{R}$ is uniformly continuous.

On the internet I found out that a function is uniformly continuous when
$\forall \varepsilon > 0 \ \exists \delta > 0: \left | f(x)-f(x_{0}) \right | < \varepsilon$ whenever $\left | x - x_{0} \right | < \delta .$
Because I don't know how to prove it calculative, I have drawn the function and showed its uniform continuity like that.
But I'd like to know how to do it the other, more professional and efficient way. I've watched some videos but anyway couldn't find a solution. Also tried to for almost 2 hours myself but nothing came out, too.
For the drawing, I think there is actually a mistake, at the beginning the epsilon (first one I haven't drawed) seems smaller than the others, while the others have same size.
(In addition I skipped the other part of the function because it's trivial).
Here is the picture:


Hint:
$f(x)=\frac{x^2}{x+1}$ is the sum between two uniformly continuous functions over $\mathbb{R}^+$: $x-1$ and $\frac{1}{x+1}$.
