Let f(x) = 4 and f’(x) = 4. Then limx→0xf(2)−2f(x)x−2 is given by
2
-2
-4
Apply L'Hospital Rule, we have,
limx→0xf(2)−2f(x)x−2(00)
limx→0f(2)−2f′(x) =f(2)−2f′(0)=4−2×4=−4