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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