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.) $$= \displaystyle{ \ e^ { \ \displaystyle{ \lim_{x \to 0^+ } \ { -2\sin x \cdot \cos x \over \cos x - x \sin x } } } }$$ $$= \displaystyle{ \ e^ { \ \displaystyle{ { -2 \sin 0 \cdot \cos 0 \over \cos 0 - 0 \cdot \sin 0 } } } }$$ $$= \displaystyle{ \ e^ { \ \displaystyle{ { -2(0)(1) \over (1) - (0)(0) } } } }$$ $$= \displaystyle{ \ e^ { \ \displaystyle{ { 0} } } }$$ $$= 1$$