SOLUTION 21: $$ \displaystyle{ \lim_{x \to 0^+} \ x^{ \ \sin x } }
= \displaystyle{ `` \ 0^{ \ \sin 0 } \ " }
= \displaystyle{ `` \ 0^{ \ 0 } \ " } $$
(Rewrite the problem to circumvent this indeterminate form. Recall that $ \displaystyle{ e^{\ln z} = z }. $)
$$ \displaystyle{ \lim_{x \to 0^+} \ x^{ \ \sin x } } = \displaystyle{ \lim_{x \to 0^+ } \ e^{ \ \ln x^{ \ \sin x } } } $$
$$ = \displaystyle{ \lim_{x \to 0^+ } \ e^{ \ { \sin x \cdot \ln x } } } $$
$$ = \displaystyle{ \ e^ { \ \displaystyle{ \lim_{x \to 0^+ } \ { \sin x \ln x }}}} = \displaystyle{ \ e^ { \ \displaystyle{ `` \ \sin 0 \cdot \ln 0 \ " } } }
= \displaystyle{ \ e^ { \ \displaystyle{ `` \ 0 \cdot (-\infty) \ " } } } $$
(``Flip" $ \sin x$ to circumvent this indeterminate form. Recall that $ \csc x = 1/\sin x $.)
$$ = \displaystyle{ \ e^ { \ \displaystyle{ \lim_{x \to 0^+ } \ { \ln x \over {1/ \sin x} } } } } $$
$$ = \displaystyle{ \ e^ { \ \displaystyle{ \lim_{x \to 0^+ } \ { \ln x \over \csc x } } } }
= \displaystyle{ \ e^ { \ \displaystyle{ `` \ \ln 0 / \csc 0 \ " } } }
= \displaystyle{ \ e^ { \ \displaystyle{ `` \ -\infty / (\cos 0/\sin 0^+) \ " } } }
= \displaystyle{ \ e^ { \ \displaystyle{ `` \ -\infty / (1/0^+) \ " } } }
= \displaystyle{ \ e^ { \ \displaystyle{ { `` -\infty / \infty \ " } } } } $$
(Apply Theorem 2 for l'Hopital's Rule. Recall that $ D\{ \csc x\} = - \csc x \cot x $ and $ \cot x = \cos x/\sin x $.)
$$ = \displaystyle{ \ e^ { \ \displaystyle{ \lim_{x \to 0^+ } \ { {1/x} \over {-\csc x \cot x} } } } } $$
$$ = \displaystyle{ \ e^ { \ \displaystyle{ \lim_{x \to 0^+ } \
{ {1/x} \over {-1 \over \sin x} \cdot { \cos x \over \sin x } } } } } $$
$$ = \displaystyle{ \ e^ { \ \displaystyle{ \lim_{x \to 0^+ } \ {1 \over x} \cdot
{ - \sin^2 x \over \cos x } } } }
= \displaystyle{ \ e^ { \ \displaystyle{ \lim_{x \to 0^+ } \
{ - \sin^2 x \over x \cos x } } } }
= \displaystyle{ \ e^ { \ \displaystyle{ `` \ -\sin^2 0 \ " \over 0 \cdot \cos 0 } } }
= \displaystyle{ \ e^ { \ \displaystyle{ `` \ -(0)^2 \ " \over (0)(1) } } }
= e^{\displaystyle{ `` \ 0/0 \ " }} $$
(Apply Theorem 1 for l'Hopital's Rule. Use the Chain Rule in the top and the Product Rule in the bottom.)
$$ = \displaystyle{ \ e^ { \ \displaystyle{ \lim_{x \to 0^+ } \
{ -2\sin x \cdot \cos x \over \cos x - x \sin x } } } } $$
$$ = \displaystyle{ \ e^ { \ \displaystyle{
{ -2 \sin 0 \cdot \cos 0 \over \cos 0 - 0 \cdot \sin 0 } } } } $$
$$ = \displaystyle{ \ e^ { \ \displaystyle{
{ -2(0)(1) \over (1) - (0)(0) } } } } $$
$$ = \displaystyle{ \ e^ { \ \displaystyle{
{ 0} } } } $$
$$ = 1 $$
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