SOLUTION 20: $$\displaystyle{ \lim_{x \to 0^+} \ { \ln x \cdot \tan x } } = \displaystyle{  \ \ln 0 \cdot \tan 0 \ " } = \displaystyle{  \ (-\infty) \cdot 0 \ " }$$ (Rewrite the expression to circumvent this indeterminate form. Recall that $\cot x = 1/\tan x$ and $\cot x = \cos x/\sin x$.) $$\displaystyle{ \lim_{x \to 0^+} \ { \ln x \cdot \tan x } } = \displaystyle{ \lim_{x \to 0^+} \ { \ln x \over 1/\tan x } }$$ $$= \displaystyle{ \lim_{x \to 0^+} \ { \ln x \over \cot x } } = \displaystyle{  \ \ln 0 \ " \over \cot 0 } = \displaystyle{  \ -\infty \ " \over \cos 0^+/\sin 0^+ } = \displaystyle{  \ -\infty \ " \over 1/0^+ } = \displaystyle{  \ - \infty \ " \over \infty }$$ (Apply Theorem 2 for l'Hopital's Rule. Recall that $D\{ \cot x\} = -\csc^2x$.) $$= \displaystyle{ \lim_{x \to 0^+} \ { 1/x \over -\csc^2 x } }$$ (Recall that $\csc x = 1/ \sin x$.) $$= \displaystyle{ \lim_{x \to 0^+} \ { 1/x \over -1/ \sin^2 x } }$$ $$= \displaystyle{ \lim_{x \to 0^+} \ { 1 \over x} \cdot { \sin^2 x \over -1 } }$$ $$= \displaystyle{ \lim_{x \to 0^+} \ { - \sin^2 x \over x } } = \displaystyle{{  \ - \sin^2 0 \ " \over 0 } } = \displaystyle{{ \ (0)^2 \ " \over 0 } } = \displaystyle{{ \ 0 \ " \over 0 } }$$ (Apply Theorem 1 for l'Hopital's Rule. Use the Chain Rule in the top.) $$= \displaystyle{ \lim_{x \to 0^+} \ { - 2 \sin x \cos x \over 1 } }$$ $$= \displaystyle{ - 2 \sin 0 \cdot \cos 0 }$$ $$= \displaystyle{ - 2 (0) (1) }$$ $$= \displaystyle{ 0 }$$