Value of $x \log x$ at $x=0$ What is the value of $f(x) = x \ln (x)$ at $x= 0$? 
Is it $0$ or indeterminate?
 A: That is an example of poor wording for a question. There is no value for the function $x\mapsto x\ln(x)$ at $x=0$ because $\ln(0)$ is not well-defined. Assuming we're talking about real and not complex logarithm, the natural domain is $(0,\infty)$ so the value of this function for $x=0$ simply does not exist. Of course one can now just define a new function using $x\ln(x)$ for $x>0$ and assign any value for $x=0$, e.g. $$\tilde f(x)=\begin{cases} x\ln(x),& x>0 \\ \pi, & x=0\end{cases}\quad\text{or}\quad \tilde f(x)=\begin{cases}x\ln(x),&x>0 \\ 42 & x=0\end{cases}\quad\text{or}~\ldots$$
thus there is not only one possible answer.
However, if one asks for the extended function to be continuous (which is a "natural" thing to do), this becomes a different question. Let $g:(a,b)\rightarrow \mathbb R$ with $a,b\in\mathbb R, a<b$ be continuous. Then $$\tilde g:[a,b)\rightarrow\mathbb R,~\begin{cases} g(x),&x\in(a,b) \\ c, & x=a\end{cases}$$ is continuous if and only if $\lim\limits_{x\downarrow a}g(x)=c$. We extend the domain of $g$ to include the boundary point $a$ and assing a special value to it. Now, if the natural domain of $g$ already includes the intervall $[a,b)$, that isn't difficult to do, but if $g(a)$ is not well-defined one has to calculate the limit for $x\downarrow a$ and see if it exists. This concept can be generalised when talking about singularities.
For your question: as $\lim\limits_{x\downarrow 0} x\ln(x)=0$ (which you should prove first if you don't know this limit) we can define $$\tilde f:[0,\infty)\rightarrow\mathbb R,~x\mapsto\begin{cases} x\ln(x),&x>0 \\ 0,&x=0\end{cases}$$ and $\tilde f$ is continuous. Thus the "natural extention" gives us the "value" of $x\ln(x)$ for $x=0$.
