SOLUTION 16: $$\displaystyle{ \lim_{x \to \infty} \ { e^x + {2/x} \over e^x + {5/x } } } = \displaystyle{  \ e^{\infty} + 2/ \infty \ " \over e^{\infty} + 5/ \infty } = \displaystyle{  \ \infty + 0 \ " \over \infty + 0 } = \displaystyle{  \ \infty \ " \over \infty }$$ (Apply Theorem 2. Differentiate the top and the bottom.) $$\displaystyle{ \lim_{x \to \infty} \ { e^x - 2/x^2 \over e^x - 5/x^2 } } = \displaystyle{  \ \infty - 0 \ " \over \infty - 0 } = \displaystyle{  \ \infty \ " \over \infty }$$ (Applying Theorem 2 for l'Hopital's Rule again will get us nowhere. Stop and think. Go back to the original problem and rewrite it.) $$\displaystyle{ \lim_{x \to \infty} \ { e^x + {2/x} \over e^x + {5/x } } } = \displaystyle{ \lim_{x \to \infty} \ { { e^x + {2/x} \over e^x + {5/x } } \cdot { x \over x } } }$$ $$= \displaystyle{ \lim_{x \to \infty} \ { xe^x + 2 \over xe^x + 5 } } = \displaystyle{  \ \infty \cdot \infty + 2 \ " \over \infty \cdot \infty + 5 } = \displaystyle{  \ \infty + 2 \ " \over \infty + 5 } = \displaystyle{  \ \infty \ " \over \infty }$$ (Apply Theorem 2 for l'Hopital's Rule. Differentiate the top and bottom using the product rule.) $$= \displaystyle{ \lim_{x \to \infty} \ { xe^x + (1)e^x + 0 \over xe^x + (1)e^x + 0} }$$ $$= \displaystyle{ \lim_{x \to \infty} \ { 1 } }$$ $$= \displaystyle{ { 1 } }$$