SOLUTION 11: $$\displaystyle{ \lim_{x \to \infty} \ { e^{3x} \over 5x+200} } = \displaystyle{  \ e^{3 \cdot \infty} \ " \over 5 \cdot \infty + 200 } = \displaystyle{  \ e^{ \infty} \ " \over \infty } = \displaystyle{  \ \infty \ " \over \infty }$$ (Apply Theorem 2 for l'Hopital's Rule. Differentiate the top using the chain rule and the bottom. Recall that $D \{ e^{f(x)} \} = e^{f(x)} \cdot f'(x)$ .) $$= \displaystyle{ \lim_{x \to \infty} \ { 3e^{3x} \over 5+0 } }$$ $$= \displaystyle{ { 3e^{\infty} \over 5 } }$$ $$= \displaystyle{ { {\infty} \over 5 } }$$ $$= \displaystyle{ \infty }$$ (So the limit does not exist.)