SOLUTION 25: $$\displaystyle{ \lim_{x \to \infty} \ x^{ \ 1/ \sqrt{x}} } = \displaystyle{  \ \infty^{ \ 1/ \infty } \ " } = \displaystyle{  \ \infty^{ \ 0} \ " }$$ (Rewrite the problem to circumvent this indeterminate form. Recall that $\displaystyle{ e^{\ln z} = z }.$) $$\displaystyle{ \lim_{x \to \infty} \ x^{ \ 1/ \sqrt{x}} } = \displaystyle{ \lim_{x \to \infty } \ e^{ \ \displaystyle \ln x^{1/ \sqrt{x}} } }$$ $$= \displaystyle{ \lim_{x \to \infty } \ e^{ \ \displaystyle (1/ \sqrt{x}) \ln x } }$$ $$= \displaystyle{ \lim_{x \to \infty } \ e^{ \ \displaystyle (\ln x) / \sqrt{x} } }$$ $$= \displaystyle{ e^{ \ \displaystyle{ \lim_{x \to \infty } \ { (\ln x) / \sqrt{x} } } } } = \displaystyle{ \ e^{ \ \displaystyle{  \ (\ln (\infty)) / \sqrt{\infty} \ " } } } = \displaystyle{ \ e^{ \ \displaystyle{  \ \infty / \infty \ " } } }$$ (Apply Theorem 2 for l'Hopital's Rule.) $$= \displaystyle{ e^{ \ \displaystyle{ \lim_{x \to \infty } \ { (1/x) / (1/2\sqrt{x}) } } } }$$ $$= \displaystyle{ e^{ \ \displaystyle{ \lim_{x \to \infty } \ { 2\sqrt{x} / x } } } }$$ $$= \displaystyle{ e^{ \ \displaystyle{ \lim_{x \to \infty } \ { 2 / \sqrt{x} } } } }$$ $$= \displaystyle{ e^{ \ \displaystyle{  \ { 2 / \sqrt{\infty} \ " } } } }$$ $$= \displaystyle{ e^{ \ \displaystyle{  \ { 2 / \infty \ " } } } }$$ $$= \displaystyle{ e^{ \ \displaystyle{ 0 } } }$$ $$= \displaystyle{ 1 }$$