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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 }
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