SOLUTION 26: $$ \displaystyle{ \lim_{x \to \infty} \ ( \ln x)^{1/x} }
= \displaystyle{ `` \ \ln(\infty)^{ \ 1/ \infty } \ " }
= \displaystyle{ `` \ \infty^{ \ 0} \ " } $$
(Rewrite the problem to circumvent this indeterminate form. Recall that $ \displaystyle{ e^{\ln z} = z }. $)
$$ \displaystyle{ \lim_{x \to \infty} \ ( \ln x)^{1/x} } = \displaystyle{ \lim_{x \to \infty } \ e^{ \ \displaystyle \ln( \ln x)^{1/x} } } $$
$$ = \displaystyle{ \lim_{x \to \infty } \ e^{ \ \displaystyle (1/x) \ln (\ln x) } } $$
$$ = \displaystyle{ \lim_{x \to \infty } \ e^{ \ \displaystyle \ln(\ln x) \Big/ x } } $$
$$ = \displaystyle{ \ e^{ \ \displaystyle{ `` \ \ln(\ln (\infty)) \Big/ \infty \ " } } }
= \displaystyle{ \ e^{ \ \displaystyle{ `` \ \ln (\infty) \Big/ \infty \ " } } }
= \displaystyle{ \ e^{ \ \displaystyle{ `` \ \infty \Big/ \infty \ " } } } $$
(Apply Theorem 2 for l'Hopital's Rule.)
$$ = \displaystyle{ e^{ \ \displaystyle{ \lim_{x \to \infty } \ {(1/\ln x) (1/x) \Big/ 1 } } } } $$
$$ = \displaystyle{ e^{ \ \displaystyle{ `` \ 1\Big/\infty \ " } } } $$
$$ = \displaystyle{ e^{ \displaystyle{ \ 0 } } } $$
$$ = \displaystyle{ 1 } $$
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