Calculus: Early Transcendentals, Chapter 4, 4.4, Section 4.4, Problem 54

You need to evaluate the limit but first you need to use the logaritmic property to convert the difference of logarithms into the logarithm of quotient:
lim_(x->1^+) (ln(x^7-1) - ln(x^5-1)) = lim_(x->1^+) ln((x^7-1)/(x^5-1))
lim_(x->1^+) ln((x^7-1)/(x^5-1)) = ln lim_(x->1^+)((x^7-1)/(x^5-1)) = ln 0/0
Since the limit is indetrminate 0/0 , you may use l'Hospital's rule:
ln lim_(x->1^+)((x^7-1)/(x^5-1)) = ln lim_(x->1^+)(7x^6)/(5x^4) = ln (7/5)
Hence, evaluating the limit of the function using l'Hospital's rule yields lim_(x->1^+) (ln(x^7-1) - ln(x^5-1)) = ln (7/5).