If 𝑓′ is continuous, 𝑓(4)=0, and 𝑓′(4)=15, evaluate lim𝑥→0[𝑓(4+3𝑥)+𝑓(4+5𝑥)]/𝑥 , limit = ?

Question

LHopitalsRule

please show the steps

Answer 1

By the Chain Rule, d/dx[f(g(x)] = f'(g(x))g'(x).

So, d/dx[f(4+3x)] = f'(4+3x) (3). Similarly, d/dx[f(4+5x)] = f'(4+5x)(5).

The limit has the form 0/0

So, by L'Hopital's Rule, the given limit is equivalent to:

lim_x→0 [ (3f'(4+3x) + 5f'(4+5x)) / 1] = (3f'(4) + 5f'(4)) / 1 = 3(15) + 5(15) = 120