SOLUTION 23: $$\displaystyle{ \lim_{x \to 0} \ (1 - x)^{1/x} } = \displaystyle{  \ (1-0)^{ 1/0^{\pm} } \ " } = \displaystyle{  \ (1-0)^{ \ \pm \infty } \ " } = \displaystyle{  \ 1^{ \ \pm \infty } \ " }$$ (Rewrite the problem to circumvent this indeterminate form. Recall that $\displaystyle{ e^{\ln z} = z }.$) $$\displaystyle{ \lim_{x \to 0} \ (1 - x)^{1/x} } = \displaystyle{ \lim_{x \to 0 } \ e^{ \ \displaystyle \ln (1 - x)^{1/x} } }$$ $$= \displaystyle{ \lim_{x \to 0 } \ e^{ \ \displaystyle {1/x} \cdot \ln (1-x) } }$$ $$= \displaystyle{ e^{ \ \displaystyle{ \lim_{x \to 0 } \ {\ln (1-x) \over x }}}} = \displaystyle{ \ e^ { \ \displaystyle{  \ { (\ln 1)/0} \ " } } } = \displaystyle{ \ e^ { \ \displaystyle{  \ { 0/0}\ " } } }$$ (Apply Theorem 1 for l'Hopital's Rule.) $$= \displaystyle{ \ e^ { \ \displaystyle{ \lim_{x \to 0} \ { { 1 \over 1-x } \cdot (-1) \over {1} } } } }$$ $$= \displaystyle{ \ e^ { \ \displaystyle{ { { -1/(1-0) } } } } }$$ $$= \displaystyle{ \ e^ { \ \displaystyle{-1} } }$$ $$= \displaystyle{ 1 \over e }$$