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.)
$$ = \displaystyle{ \ e^ { \ \displaystyle{ \lim_{x \to 0} \
{ { 1 \over 1-x } \cdot (-1) \over {1} } } } } $$
$$ = \displaystyle{ \ e^ { \ \displaystyle{ { { -1/(1-0) } } } } } $$
$$ = \displaystyle{ \ e^ { \ \displaystyle{-1} } } $$
$$ = \displaystyle{ 1 \over e } $$
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