I have to decide if the function $x \ln x$ is uniformly continuous in $(0,3]$.
I could show that $x \ln x$ is defined and continuous in a closed and limited interval. In this situation for Weierstrass I could prove that $x \ln x $ is uniformly continuous in such interval.
And then use the extension theorem for which $x \ln x$ is uniformly continuous also in a restriction such as $(0,3]$.
But $x \ln x$ is not defined in 0. Could someone help me to understand?
$\lim_{x \to 0+} x \ln x =0$ so $f(x)=x \ln x$ for $0 <x\leq 3, f(0)=0$ defines a continuous function on $[0,3]$. Hence $f$ is uniformly continuous and this implies that $x \ln x $ is uniformly continuous $(0,3]$.
To prove that $x \ln x \to 0$ as $x \to 0+$ apply L'Hopital's Rule to $\frac {\ln x} {1/x}$.