SOLUTION 19: $$ \displaystyle{ \lim_{x \to 0^+} \ { x \cdot ( \ln x )^2 } }
= `` \ 0 \cdot ( \ln 0 )^2 \ " = `` \ 0 \cdot (-\infty)^2 \ " = `` \ 0 \cdot \infty \ " $$
(This is an indeterminate form. ``Flip" $x$ so that l'Hopital's Rule can be applied.)
$$ \displaystyle{ \lim_{x \to 0^+} \ { x \cdot ( \ln x )^2 } }
= \displaystyle{ \lim_{x \to 0^+} \ { (\ln x)^2 \over 1/x } }
= \displaystyle{ { {`` \ (\ln 0)^2 \ "} \over 1/0^+ } }
= \displaystyle{ `` \ (\infty)^2 \ " \over \infty }
= \displaystyle{ `` \ \infty \ " \over \infty } $$
(Apply Theorem 2 for l'Hopital's Rule.)
$$ = \displaystyle{ \lim_{x \to 0^+} \ {2 \ln x \cdot 1/x \over -1/x^2} } $$
$$ = \displaystyle{ \lim_{x \to 0^+} \ {2 \ln x \over -1/x} }
= \displaystyle{ `` \ 2 \ln 0 \ " \over {-1/0^+} }
= \displaystyle{ `` \ 2(-\infty) \ " \over -\infty }
= \displaystyle{ `` \ -\infty \ " \over -\infty } $$
(Apply Theorem 2 for l'Hopital's Rule again.)
$$ = \displaystyle{ \lim_{x \to 0^+} \ { { 2 \cdot 1/x } \over { 1/x^2 } } } $$
$$ = \displaystyle{ \lim_{x \to 0^+ } \ 2x } $$
$$ = 2 \cdot 0 $$
$$ = 0 $$
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