SOLUTION 14: $$\displaystyle{ \lim_{x \to \infty} \ { (\ln x)^2 \over e^{2x} } } = \displaystyle{  \ (\ln(\infty))^2 \ " \over e^{2\infty} } = \displaystyle{  \ (\infty)^2 \ " \over e^{\infty} } = \displaystyle{  \ \infty \ " \over \infty }$$ (Apply Theorem 2 for l'Hopital's Rule. Differentiate the top and the bottom using the chain rule. Recall that $D \{ e^{f(x)} \} = e^{f(x)} \cdot f'(x)$ .) $$= \displaystyle{ \lim_{x \to \infty} \ { 2(\ln x) \cdot (1/x) \over 2e^{2x} } }$$ $$= \displaystyle{ \lim_{x \to \infty} \ { \ln x \over xe^{2x} } } = \displaystyle{  \ \ln(\infty) \ " \over (\infty)e^{2\infty} } = \displaystyle{  \ \infty \ " \over (\infty)e^{\infty} } = \displaystyle{  \ \infty \ " \over (\infty)(\infty) } = \displaystyle{  \ \infty \ " \over \infty }$$ (Apply Theorem 2 for l'Hopital's Rule again. Differentiate the bottom using the product rule and the chain rule.) $$= \displaystyle{ \lim_{x \to \infty} \ { 1/x \over x \cdot 2e^{2x}+(1)e^{2x} } }$$ $$= \displaystyle{ \lim_{x \to \infty} \ { {1 \over x} \cdot { 1 \over 2xe^{2x}+ e^{2x} } } }$$ $$= \displaystyle{ \lim_{x \to \infty} \ { {1 \over 2x^2e^{2x}+xe^{2x} } } }$$ $$= \displaystyle{ 1 \over 2(\infty)^2e^{2\infty}+(\infty)e^{2\infty} }$$ $$= \displaystyle{ 1 \over \infty + \infty }$$ $$= \displaystyle{ 1 \over \infty }$$ $$= \displaystyle{ 0 }$$