1) We'll substitute x by the value 1 and we'll calcualte the limit.

lim (x^a - 1)/(x^b - 1) = (1^a - 1)/(1^b - 1) = (1-1)/(1-1) = 0/0

Since the result represents an indetermination, we'll apply L'Hospital rule.

lim (x^a - 1)/(x^b - 1) = lim (x^a - 1)'/(x^b - 1)'

(x^a - 1)' = a*x^(a-1)

(x^b - 1)' = b*x^(b-1)

lim (x^a - 1)'/(x^b - 1)' = lim a*x^(a-1)/b*x^(b-1)

We'll apply the quotient rule of the power function:

x^m/x^n = x^(m-n)

We'll put m = a-1 and n = b-1

x^(a-1)/x^(b-1) = x^(a-1-b+1)

x^(a-1)/x^(b-1) = x^(a-b)

lim a*x^(a-1)/b*x^(b-1) = (a/b)*lim x^(a-b)

We'll substitute x by 1:

(a/b)*lim x^(a-b) = (a/b)*1^(a-b) = a/b

**lim (x^a - 1)/(x^b - 1) = a/b**

1

lt x-->1 (x^a-1)/(x^b-1).

We know that (x^n-a^n)/(x-a) = x^n-1+x^n-2*a+x^n-2*a^2+...xa^n-2+a^n-1.

Therefore Lt x->a (x^n-a^n)/(x-a) = n a^n-1.

Therfore Lt x-->1 (x^a-1)/(x^b-1) = Lt {(x^a-1)/(x-1)}/{x^b -1)/(x-1)} = a*1^(a-1)/b*1^(a-1) = a/b.