Use L'Hopitals Rule to find the following: (x^a) - 1 / (x^b) - 1?

Question

Using L'Hopital's Rule f'(x)/g'(x) I got...

ax^(a-1) / bx^(b-1) but that's not correct. How do I go about getting the correct answer?

Answer 1

lim [x^a - 1] / [x^b - 1]
x -> 1

Plugging in x = 1, we would have (1^a - 1) / (1^b - 1) =
(1 - 1)/(1 - 1) = [0/0], so we can freely use L'Hospital's rule. Doing so, we obtain

lim [ax^(a - 1)] / [bx^(b - 1)]
x -> 1

Pulling the constant (a/b) out of the limit, we have

(a/b) lim [x^(a - 1) / x^(b - 1)]
. . . . x -> 1

Safely plugging in x = 1, we have

(a/b) (1/1)

or just

(a/b)

You forgot to take it a step further and calculate the limit as x approaches 1.

Answer 2

(f.g) ' = f'.g + g'f

for you :
f = x^1-1
g = 1/ (x^b-1)