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.) $$= \displaystyle{ e^{ \ \displaystyle{ \lim_{x \to \infty } \ {(1/\ln x) (1/x) \Big/ 1 } } } }$$ $$= \displaystyle{ e^{ \ \displaystyle{  \ 1\Big/\infty \ " } } }$$ $$= \displaystyle{ e^{ \displaystyle{ \ 0 } } }$$ $$= \displaystyle{ 1 }$$