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.) $$= \displaystyle{ \ e^ { \ \displaystyle{ \lim_{x \to 0^+} \ { { -x^3 \cdot 2 \sec x \cdot \sec x \tan x -3x^2 \sec^2 x \over 2 \sec^2 x }}}}}$$ $$= \displaystyle{ \ e^ { \ \displaystyle{ \lim_{x \to 0^+} \ { { -2x^3 \sec^2 x \tan x -3x^2 \sec^2 x \over 2 \sec^2 x }}}}}$$ $$= \displaystyle{ \ e^ { \ \displaystyle{ { { 0 \cdot \sec^2 0 \cdot \tan 0 - 0 \cdot \sec^2 0 \over 2 \sec^2 0 }}}}}$$ $$= \displaystyle{ \ e^ { \ \displaystyle{ { { (0)(1)^2 (0) - (0)(1)^2 \over 2 (1)^2 }}}}}$$ $$= \displaystyle{ \ e^ { \ \displaystyle{0/2} }}$$ $$= e^{ \ \displaystyle{0}}$$ $$= 1$$