SOLUTION 13: $$ \displaystyle{ \lim_{x \to \infty} \ {x^2+3x-10 \over 7x^2-5x+4} } = \displaystyle{ \lim_{x \to \infty} \ {x(x+3)-10 \over x(7x-5)+4} } = \displaystyle{ `` \ (\infty)(\infty)-10 \ " \over (\infty)(\infty)+4 } = \displaystyle{ `` \ \infty-10 \ " \over \infty+4 } = \displaystyle{ `` \ \infty \ " \over \infty } $$
(Apply Theorem 2 for l'Hopital's Rule. Differentiate the top and the bottom.)
$$ = \displaystyle{ \lim_{x \to \infty} \ { 2x+3-0 \over 14x-5+0 } } = \displaystyle{ `` \ 2(\infty)+3 \ " \over 14(\infty)-5 }
= \displaystyle{ `` \ \infty+3 \ " \over \infty-5 } = \displaystyle{ `` \ \infty \ " \over \infty } $$
(Apply Theorem 2 for l'Hopital's Rule again.)
$$ = \displaystyle{ \lim_{x \to \infty} \ {2+0 \over 14-0} } $$
$$ = \displaystyle{ 1 \over 7} $$
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