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.) $$= \displaystyle{ \lim_{x \to \infty} \ {2+0 \over 14-0} }$$ $$= \displaystyle{ 1 \over 7}$$