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 $ )
$$ = \displaystyle{ (0) { (\ln 2)^2 \over (\ln 3 )^2 } } $$
$$ = \displaystyle{ 0 } $$
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