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 }. $) $$ \displaystyle{ \lim_{x \to \infty} \ ( \ln x)^{1/x} } = \displaystyle{ \lim_{x \to \infty } \ e^{ \ \displaystyle \ln x^{1/x} } } $$ $$ = \displaystyle{ \lim_{x \to \infty } \ e^{ \ \displaystyle (1/x) \ln x } } $$ $$ = \displaystyle{ \lim_{x \to \infty } \ e^{ \ \displaystyle (\ln x) / x } } $$ $$ = \displaystyle{ \ e^{ \lim_{x \to \infty } \ \displaystyle (\ln x) / x } } = \displaystyle{ \ e^{ \ \displaystyle{ `` \ (\ln (\infty)) / \sqrt{\infty} \ " } } } = \displaystyle{ \ e^{ \ \displaystyle{ `` \ \infty / \infty \ " } } } $$ (Apply Theorem 2 for l'Hopital's Rule.) $$ = \displaystyle{ e^{ \ \displaystyle{ \lim_{x \to \infty } \ { (1/x) / 1 } } } } $$ $$ = \displaystyle{ e^{ \ \displaystyle{ `` \ 1/\infty \ " } } } $$ $$ = \displaystyle{ e^{ \displaystyle{ \ 0 } } } $$ $$ = \displaystyle{ 1 } $$

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