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.)
$$ = \displaystyle{ \lim_{x \to 0^+} \ { 1/x \over -1/x^2} } $$
$$ = \displaystyle{ \lim_{x \to 0^+} \ { 1 \over x } \cdot { x^2 \over -1 } } $$
$$ =\displaystyle{ \lim_{x \to 0^+ } \ (-x) } $$
$$ = 0 $$
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