Let f(x) = 4 and f′(x) = 4

• Limits Continuity

Let f(x) = 4 and f′(x) = 4. Then Lim_x→2 (xf(2) - 2f(x))/(x-2) is given by

1. -4
2. 2
3. -2
4. 3

Answer

Lim_x→2 (xf(2) - 2f(2))/(x-2) can be written as

Lim_x→2 {f(2) - 2f'(2)}  [L' Hospitals Rule]

f(2) - 2f'(2)

= 4 - 2x4

= -4

The correct option is A.