## Calculus: Early Transcendentals 8th Edition

$\lim\limits_{x \to 0}\frac{f(2+3x)+f(2+5x)}{x} = 56$
$\lim\limits_{x \to 0}\frac{f(2+3x)+f(2+5x)}{x} = \frac{0+0}{0} = \frac{0}{0}$ We can apply L'Hospital's Rule. $\lim\limits_{x \to 0}\frac{3f'(2+3x)+5f'(2+5x)}{1} = 3(7)+5(7) = 56$ Therefore: $\lim\limits_{x \to 0}\frac{f(2+3x)+f(2+5x)}{x} = 56$