SOLUTION 27: $$ \displaystyle{ \lim_{x \to 0^+} \ x^{x^x} }
= \displaystyle{ `` \ 0^{ \ 0^{ \ 0}} \ " } $$
(Rewrite the problem to circumvent this indeterminate form. Recall that $ \displaystyle{ e^{\ln z} = z }. $)
$$ \displaystyle{ \lim_{x \to 0^+} \ x^{x^x} } = \displaystyle{ \lim_{x \to 0^+ } \ e^{ \ \displaystyle \ln x^{x^x} } } $$
$$ = \displaystyle{ \lim_{x \to 0^+ } \ e^{ \ \displaystyle x^x \ln x } } $$
$$ = \displaystyle{ e^{ \ \displaystyle{ \lim_{x \to 0^+ } \ x^x \ln x }}} $$
$$ = \displaystyle{ e^{ \
\displaystyle{ (\lim_{x \to 0^+ } \ x^x)( \lim_{x \to 0^+ } \ \ln x) }}} $$
(See Example 4 in the material preceeding this problem set.)
$$ = \displaystyle{ e^{ \ \displaystyle{ `` \ (1)(\ln 0) \ " }}} $$
$$ = \displaystyle{ e^{ \ \displaystyle{ `` \ (1)(-\infty) \ " }}} $$
$$ = \displaystyle{ `` \ e^{- \infty} \ " } $$
$$ = \displaystyle{ `` \ 1/ e^{\infty} \ " } $$
$$ = \displaystyle{ `` \ 1/ \infty \ " } $$
$$ = \displaystyle{ 0 } $$
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