What is lim h->0 ((8 (1/2+h)^8) - 8 (1/2)^8)/h ?

Answer 1

you could just use L'Hopital's rule

lim h->0 ((8 (1/2+h)^8) - 8 (1/2)^8)/h
= lim 64(1/2 + h)^7 / 1 , h ---> 0
= 64(1/2)^7
= 64/128
= 1/2

Answer 2

or, use the binomial expansion ...
(8 (1/2+h)^8) - 8 (1/2)^8)
= 8[ 1/2+h)^8) - (1/2)^8 ]
= 8[ (1/2)^8 + 8(1/2)^7 h + ... + h^8 - (1/2)^8 ]
= 8[ 8(1/2)^7 h + ... + h^8 ]
= 8[ h(8(1/2)^7 + ... + h^7 ]

then lim h->0 ((8 (1/2+h)^8) - 8 (1/2)^8)/h
= lim h --> 0 ( 8[ h(8(1/2)^7 + ... + h^7 ] )/h
= lim h --> 0 ( 8[ (8(1/2)^7 + ... + h^7 ] )
= 8[8(1/128)]
= 64/128
= 1/2

Answer 3

I expect the second solution is the one intended here. It looks like a problem before calculus.

Answer 4

or,

they might want you to recognize that this is the
First Principles approach to finding the derivative of
y = 8x^8 at x = 1/2

dy/dx = 64x^7 = 64(1/2)^7 = 1/2

assuming that they are at the stage when they know how to differentiate 8x^8