Uniform continuity of $x\log(x)$ The function $f(x)=x\log(x)$ is clearly continuous on $[0,1]$ and thus uniformly continuous. 
I would like to ask whether one can explicitly state a $\delta(\varepsilon)$ such that for all $x,y \in [0,1]$
$$\left\lvert x-y \right\rvert \le \delta \Rightarrow \left\lvert f(x)-f(y) \right\rvert< \varepsilon?$$
 A: For the direct $\epsilon-\delta$ approach, assuming WLOG $x > y > 0$ we have
$$|x \log x - y \log y| \leqslant |x-y| |\log x| + |y| |\log x - \log y|\\ = |\sqrt{x}- \sqrt{y}||\sqrt{x}+\sqrt{y}||2 \log \sqrt{x}| +  |y| |\log \left(1 + \frac{x-y}{y} \right)$$
Applying the inequalities $|\log s| < (1-s)/s$ for $s \in (0,1]$ and $\log(1+t) < t$ for $t > 0$ with $s = \sqrt{x}$ and $t = (x-y)/y$ we get 
$$\tag{*}|x \log x - y \log y| \leqslant \frac{2|\sqrt{x} + \sqrt{y}|}{\sqrt{x}}(1- \sqrt{x})|\sqrt{x} - \sqrt{y}| + |y|\frac{|x-y|}{|y|}\\ \frac{4\sqrt{x}}{\sqrt{x}}(1- \sqrt{x})|\sqrt{x} - \sqrt{y}| + |y|\frac{|x-y|}{|y|}\\ \leqslant 4|\sqrt{x} - \sqrt{y}| + |x-y|$$
Since, $x \mapsto \sqrt{x}$ and $x \mapsto x$ are uniformly continuous on $[0,1]$, we can find $\delta(\epsilon) > 0$ such that the RHS of (*) is less than $\epsilon$ when $|x-y| < \delta(\epsilon)$ for all $x,y \in (0,1]$.
If the function $f$ is defined such that $f(0)= 0$, then this can be handled using $\lim_{x \to 0} z \log z = 0$, extending uniform continuity to $[0,1]$.
