Answer
$\lim\limits_{x \to 0}\frac{f(2+3x)+f(2+5x)}{x} = 56$
Work Step by Step
$\lim\limits_{x \to 0}\frac{f(2+3x)+f(2+5x)}{x} = \frac{0+0}{0} = \frac{0}{0}$
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:
$\lim\limits_{x \to 0}\frac{f(2+3x)+f(2+5x)}{x} = 56$