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.)
$$ = \displaystyle{ e^{ \ \displaystyle{ \lim_{x \to \infty } \ { (1/x) / (1/2\sqrt{x}) } } } } $$
$$ = \displaystyle{ e^{ \ \displaystyle{ \lim_{x \to \infty } \ { 2\sqrt{x} / x } } } } $$
$$ = \displaystyle{ e^{ \ \displaystyle{ \lim_{x \to \infty } \ { 2 / \sqrt{x} } } } } $$
$$ = \displaystyle{ e^{ \ \displaystyle{ `` \ { 2 / \sqrt{\infty} \ " } } } } $$
$$ = \displaystyle{ e^{ \ \displaystyle{ `` \ { 2 / \infty \ " } } } } $$
$$ = \displaystyle{ e^{ \ \displaystyle{ 0 } } } $$
$$ = \displaystyle{ 1 } $$
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