How to evaluate the $\lim_{x\to a-}(a-x)\ln(a-x)$? I know it ought to be zero but I don't know the rule to evaluate this limit:
$$\lim_{x\to a^{-}}(a-x)\ln(a-x)$$
Any Ideas?
 A: Sketch: Typically one does $$\lim_{x\to a^-}(a-x)\ln(a-x)=\lim_{x\to a^-}\frac{\ln(a-x)}{\frac1{a-x}}\stackrel{\color{blue}{\text{l'Hôpital}}}= \lim_{x\to a^-}\ \frac{\color{blue}?}{\color{blue}?}$$
A: You can interpret it as a quotient and use L'Hôpital:
$$(a-x)\log(a-x) = \frac{\log(a-x)}{(a-x)^{-1}}$$
Differentiate to get:
$$\frac{-1}{(a-x)^{-1}} = -(a-x).$$
As this goes to zero you are done.
A: Apply  L'Hôpital's rule : 

$$\lim _{ x\to a- } \left( a-x \right) \ln { \left( a-x \right)  } =\lim _{ x\rightarrow { a }- }{ \frac { \ln { \left( a-x \right)  }  }{ \frac { 1 }{ \left( a-x \right)  }  }  } =\lim _{ x\rightarrow { a }- }{ \frac { -\frac { 1 }{ a-x }  }{ \frac { 1 }{ { \left( a-x \right)  }^{ 2 } }  }  } =\lim _{ x\rightarrow a- }{ -\left( a-x \right)  } =0$$

A: Use the sub $u = a-x$ so that $$\lim_{x\to a^{-}} (a-x)\ln(a-x) = \lim_{u\to 0^{+}} u \ln u$$ then $u = e^{-y}$ gives $$\lim_{u \to 0^+} u\ln u = -\lim_{y \to \infty} ye^{-y}$$ since $e^y > y^2/2!$ for $y>0$ we obtain the required limit via squeezing. 
