The limit below represents a derivative f′(a). Find f(x) and a. lim h--> 0 ((6+h)^3)-216/h. a=___.

Answer 1

f'(a) = lim_h→0 [(f(a+h) - f(a)) / h]

So, if we let f(x) = x³ and a = 6. Then f(6) = 216.

Therefore, f'(6) = lim_h→0 [(f(6+h) - f(6))/h] = lim_h→0 [((6+h)³ - 216) / h] = 3(6)² = 108_.

Answer 2

Genesis,

First expand (6+h)³ = 6³+3*36h+3*6*h²+h³ = 216 + 108h +18h² +h³ and substitute. You can cancel 216 then

and divide each term by h:

lim h->0 [ ( 216 + 108h +18h² +h³ - 216) / h] = lim h->0 ( 108 +18h +h² ) = 108