SOLUTION 12: $$ \displaystyle{ \lim_{x \to \infty} \ { 3 + \ln x \over x^2+7 } } = \displaystyle{ `` \ 3+\ln(\infty) \ " \over (\infty)^2+7 } = \displaystyle{ `` \ 3+ \infty \ " \over \infty + 7 } = \displaystyle{ `` \ \infty \ " \over \infty } $$ (Apply Theorem 2 for l'Hopital's Rule. Differentiate the top and the bottom.) $$ = \displaystyle{ \lim_{x \to \infty} \ { 0+1/x \over 2x+0 } } $$ $$ = \displaystyle{ \lim_{x \to \infty} \ {1 \over x} \cdot {1 \over 2x} } $$ $$ = \displaystyle{ \lim_{x \to \infty} \ {1 \over 2x^2 } } $$ $$ = \displaystyle{ { `` \ 1 \ " \over {\infty} } } $$ $$ = \displaystyle{0} $$

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