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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