SOLUTION 10: $$\displaystyle{ \lim_{x \to 0} \ { e^{-1/x^2} \over x^2 } } = \displaystyle{  \ e^{-1/0^2} \ " \over (0)^2 } = \displaystyle{  \ e^{-\infty} \ " \over 0 } = \displaystyle{  \ { 1 / e^{\infty} } \ " \over 0 } = \displaystyle{  \ { 1 / \infty } \ " \over 0 } = \displaystyle{  \ 0 \ " \over 0 }$$ (Apply Theorem 1 for l'Hopital's Rule. Differentiate the top using the chain rule and the bottom. Recall that $D \{ e^{f(x)} \} = e^{f(x)} \cdot f'(x)$ .) $$= \displaystyle{ \lim_{x \to 0} \ { e^{-1/x^2} \cdot 2/x^3 \over 2x } }$$ $$= \displaystyle{ \lim_{x \to 0} \ { 2e^{-1/x^2} \over x^3} \cdot { 1 \over 2x } }$$ $$= \displaystyle{ \lim_{x \to 0} \ { e^{-1/x^2} \over x^4 } } = \displaystyle{  \ 0 \ " \over 0 }$$ (Applying Theorem 1 for l'Hopital's Rule again leads to $\displaystyle{ \lim_{x \to 0} \ { e^{-1/x^2} \over 2x^6 } }$. Stop and think. This is getting nowhere. Go back to the beginning and algebraically rewrite the problem by flipping" both numerator and denominator.) $$= \displaystyle{ \lim_{x \to 0} \ { e^{-1/x^2} \over x^2 } }$$ $$= \displaystyle{ \lim_{x \to 0} \ { 1/x^2 \over e^{1/x^2} } } = \displaystyle{ {  \ 1/0^2 \ " \over e^{1/0^2} } } = \displaystyle{ {  \ \infty \ " \over e^{\infty} } } = \displaystyle{  \ \infty \ " \over \infty }$$ (Apply Theorem 2 for l'Hopital's Rule.) $$= \displaystyle{ \lim_{x \to 0} \ { -2/x^3 \over e^{1/x^2} \cdot -2/x^3 } }$$ $$= \displaystyle{ \lim_{x \to 0} \ { { 1 \over e^{1/x^2} } } }$$ $$= \displaystyle{ \lim_{x \to 0} \ { { 1 \over e^{1/(0)^2} } } }$$ $$= \displaystyle{ \lim_{x \to 0} \ { { 1 \over e^{\infty} } } }$$ $$= \displaystyle{  \ 1 \ " \over \infty }$$ $$= 0$$