SOLUTION 15: $$\displaystyle{ \lim_{x \to \infty} \ { 3x+2^x \over 2x+3^x } } = \displaystyle{  \ 3(\infty)+2^{\infty} \ " \over 2(\infty)+3^{\infty} } = \displaystyle{  \ \infty + \infty \ " \over \infty + \infty } = \displaystyle{  \ \infty \ " \over \infty }$$ (Apply Theorem 2 for l'Hopital's Rule. Differentiate the top and the bottom. Recall that $D \{ a^{x)} \} = a^{x} \cdot \ln a$ .) $$= \displaystyle{ \lim_{x \to \infty} \ { 3+2^x \ln 2 \over 2+3^{x} \ln 3 } } = \displaystyle{  \ 3+2^{\infty} \ln 2 \ " \over 2+3^{\infty} \ln 3 } = \displaystyle{  \ 3 + \infty \ " \over 2 + \infty } = \displaystyle{  \ \infty \ " \over \infty }$$ (Apply Theorem 2 for l'Hopital's Rule again.) $$= \displaystyle{ \lim_{x \to \infty} \ {0+2^{x} \ln 2 \cdot \ln 2 \over 0+3^{x} \ln 3 \cdot \ln 3 } }$$ $$= \displaystyle{ \lim_{x \to \infty} \ {2^{x} (\ln 2)^2 \over 3^{x} (\ln 3)^2 } } = \displaystyle{ {2^{\infty} (\ln 2)^2 \over 3^{\infty} (\ln 3 )^2 } } = \displaystyle{ {(\infty) (\ln 2)^2 \over (\infty) (\ln 3 )^2 } } = \displaystyle{  \ \infty \ " \over \infty }$$ (Applying Theorem 2 for l'Hopital's Rule again will lead to $\displaystyle{ \lim_{x \to \infty} \ {2^{x} (\ln 2)^3 \over 3^{x} (\ln 3)^3 } }$. Stop and think. This is going nowhere. There is an easy way out.) $$= \displaystyle{ \lim_{x \to \infty} \ {2^{x} (\ln 2)^2 \over 3^{x} (\ln 3)^2 } }$$ $$= \displaystyle{ \lim_{x \to \infty} \ {2^{x} \over 3^{x} } \cdot { (\ln 2)^2 \over (\ln 3 )^2 } }$$ (Recall that $\displaystyle{a^x \over b^x} = \displaystyle{ \Big( {a \over b} \Big)^x }$.) $$= \displaystyle{ \lim_{x \to \infty} \ \Big( {2 \over 3} \Big)^x \cdot { (\ln 2)^2 \over (\ln 3 )^2 } }$$ $$= \displaystyle{ \Big( {2 \over 3} \Big)^{\infty} \cdot { (\ln 2)^2 \over (\ln 3 )^2 } }$$ (Since $-1 < \frac{2}{3} < +1$ ) $$= \displaystyle{ (0) { (\ln 2)^2 \over (\ln 3 )^2 } }$$ $$= \displaystyle{ 0 }$$