SOLUTION 24: $$ \displaystyle{ \lim_{x \to 0^+} \ (\tan x)^{x^2} }
= \displaystyle{ `` \ (\tan 0)^{ \ 0^2 } \ " }
= \displaystyle{ `` \ 0^{ \ 0 } \ " } $$
(Rewrite the problem to circumvent this indeterminate form. Recall that $ \displaystyle{ e^{\ln z} = z }. $)
$$ \displaystyle{ \lim_{x \to 0^+} \ (\tan x)^{x^2} } = \displaystyle{ \lim_{x \to 0^+ } \ e^{ \ \displaystyle \ln (\tan x)^{x^2} } } $$
$$ = \displaystyle{ \lim_{x \to 0^+ } \ e^{ \ \displaystyle x^2 \cdot \ln (\tan x) } } $$
$$ = \displaystyle{ e^{ \ \displaystyle{ \lim_{x \to 0^+ } \ x^2 \cdot \ln (\tan x) } } }
= \displaystyle{ \ e^ { \ \displaystyle{ `` \ 0 \cdot \ln (\tan 0) \ " } } }
= \displaystyle{ \ e^ { \ \displaystyle{ `` \ 0 \cdot \ln 0 \ " } } }
= \displaystyle{ \ e^ { \ \displaystyle{ `` \ 0 \cdot (-\infty) \ " } } } $$
(``Flip" $x^2$ to circumvent this indeterminate form.)
$$ = \displaystyle{ e^{ \ \displaystyle{ \lim_{x \to 0^+ } \ { \ln (\tan x) \over 1/x^2 } } } }
= \displaystyle{ \ e^ { \ \displaystyle{ `` \ \ln (\tan 0)/(1/0^2) \ " } } }
= \displaystyle{ \ e^ { \ \displaystyle{ `` \ \ln 0 / \infty \ " } } }
= \displaystyle{ \ e^ { \ \displaystyle{ `` \ -\infty / \infty \ " } } } $$
(Apply Theorem 2 for l'Hopital's Rule. Recall that $ D \{ \ln(f(x) \} = \displaystyle{ 1 \over f(x) } \cdot f'(x) $ and $ D\{ \tan x \}= \sec^2x $.)
$$ = \displaystyle{ \ e^ { \ \displaystyle{ \lim_{x \to 0^+} \
{ { (1/\tan x) } \cdot \sec^2 x \over -2/x^3 } } } } $$
$$ = \displaystyle{ \ e^ { \ \displaystyle{ \lim_{x \to 0^+} \
{ { \sec^2 x \over \tan x } \cdot {x^3 \over -2} } } } } $$
$$ = \displaystyle{ \ e^ { \ \displaystyle{ \lim_{x \to 0^+} \
{ { -x^3 \sec^2 x \over 2 \tan x } } } } }
= \displaystyle{ \ e^ { \ \displaystyle{ { { `` \ 0 \cdot \sec^2 0 \ " \over 2 \tan 0 } } } } }
= \displaystyle{ \ e^ { \ \displaystyle{ { { `` \ (0)(1)^2 \ " \over 2 (0) } } } } }
= \displaystyle{ \ e^ { \ \displaystyle{ `` \ 0/0 \ " } } } $$
(Apply Theorem 1 for l'Hopital's Rule.)
$$ = \displaystyle{ \ e^ { \ \displaystyle{ \lim_{x \to 0^+} \
{ { -x^3 \cdot 2 \sec x \cdot \sec x \tan x -3x^2 \sec^2 x \over 2 \sec^2 x }}}}} $$
$$ = \displaystyle{ \ e^ { \ \displaystyle{ \lim_{x \to 0^+} \
{ { -2x^3 \sec^2 x \tan x -3x^2 \sec^2 x \over 2 \sec^2 x }}}}} $$
$$ = \displaystyle{ \ e^ { \ \displaystyle{
{ { 0 \cdot \sec^2 0 \cdot \tan 0 - 0 \cdot \sec^2 0 \over 2 \sec^2 0 }}}}} $$
$$ = \displaystyle{ \ e^ { \ \displaystyle{
{ { (0)(1)^2 (0) - (0)(1)^2 \over 2 (1)^2 }}}}} $$
$$ = \displaystyle{ \ e^ { \ \displaystyle{0/2} }} $$
$$ = e^{ \ \displaystyle{0}} $$
$$ = 1 $$
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