How to solve lim as x->1 where f(x) = (x^7-1)/(x^92-1) WITHOUT l'hopitals rule

Hint:
1+x+...+x^6 = (x^7 - 1)/(x-1)
1+x+...+x^91 = (x^92 - 1)/(x-1)
So,
lim{x->1} (x^7-1)/(x^92-1)
= lim{x->1} (1+x+...+x^6)/(1+x+...+x^91)
= 7/92