SOLUTION 9: $$\displaystyle{ \lim_{x \to 0} \ { x^2 e^x \over \tan^2 x} } = \displaystyle{  \ 0 \cdot e^{0} \ " \over \tan^2 0 } = \displaystyle{  \ 0 \cdot 1 \ " \over (0)^2 } = \displaystyle{  \ 0 \ " \over 0 }$$ (Apply Theorem 1 for l'Hopital's Rule. Differentiate the top using the product rule and the bottom using the chain rule.) $$= \displaystyle{ \lim_{x \to 0} \ {x^2 e^x + 2xe^x \over 2 \tan x \cdot \sec^2x} } = \displaystyle{ {(0)^2 e^{0} + 2(0) e^{0} \over 2 \tan 0 \cdot \sec^2 0 } } = \displaystyle{ {(0)(1)+(0)(1) \over (0)(1)^2 } } = \displaystyle{  \ 0 \ " \over 0 }$$ (Apply Theorem 1 for l'Hopital's Rule again. Differentiate the top using the product rule and the bottom using the product rule and the chain rule.) $$= \displaystyle{ \lim_{x \to 0} \ {(x^2e^x+2xe^x)+(2xe^x+2e^x) \over 2 \tan x \cdot 2 \sec x \cdot \sec x \tan x + 2 \sec^2 x \cdot \sec^2 x } }$$ $$= \displaystyle{ \lim_{x \to 0} \ {x^2e^x+4xe^x+2e^x \over 4 \tan^2 x \cdot \sec^2 x + 2 \sec^4 x } }$$ $$= \displaystyle{ {(0)^2 e^{0} + 4(0) e^{0}+2e^{0} \over 4 \tan^2 0 \cdot \sec^2 0 + 2\sec^4 0 } }$$ $$= \displaystyle{ {(0)(1)+(0)(1)+2(1) \over 4(0)^2 (1)^2+2(1)^4 } }$$ $$= \displaystyle{ 2 \over 2 }$$ $$= 1$$