SOLUTION 18: $$\displaystyle{ \lim_{x \to 0^+} \ { x \cdot \ln x} } =  \ 0 \cdot \ln 0 \ " =  \ 0 \cdot (-\infty) \ "$$ (This is an indeterminate form. Flip" $x$ so that l'Hopital's Rule can be applied.) $$\displaystyle{ \lim_{x \to 0^+} \ { x \cdot \ln x} } = \displaystyle{ \lim_{x \to 0^+} \ { \ln x \over 1/x } } = \displaystyle{ {  \ \ln 0 \ " \over 1/0^+ } } = \displaystyle{  \ - \infty \ " \over \infty }$$ (Apply Theorem 2 for l'Hopital's Rule. Differentiate the top and the bottom separately.) $$= \displaystyle{ \lim_{x \to 0^+} \ { 1/x \over -1/x^2} }$$ $$= \displaystyle{ \lim_{x \to 0^+} \ { 1 \over x } \cdot { x^2 \over -1 } }$$ $$=\displaystyle{ \lim_{x \to 0^+ } \ (-x) }$$ $$= 0$$