# How to evaluate lim x->1 [1/(1-x) – 3/(1-x^3)] and lim x->1 (x^n-1)/(x-1)?

##lim_(x->1) [1/(1-x) – 3/(1-x^3)] = -1## and ##lim_(x->1) (x^n-1)/(x-1) = n##

First Limit
##lim_(x->1) [1/(1-x) – 3/(1-x^3)]##
We cannot use the for limits, because the denominator has a limit of ##0##.
Write as a single ratio.
I will use ##(1-x^3) = (1-x)(1+x+x^2)## (difference of cubes).
##lim_(x->1) [1/(1-x) – 3/(1-x^3)] = lim_(x->1) [(1+x+x^2)/((1-x)(1+x+x^2)) – 3/(1-x^3)]##

## = lim_(x->1) ((1+x+x^2) – 3)/(1-x^3)##
## = lim_(x->1) [-(x^2+x-2)/(x^3-1)]##
## = lim_(x->1) [-((x+2)(x-1))/((x-1)(x^2+x+1))]##
## = lim_(x->1) [-(x+2)/(x^2+x+1)]##
## = -3/3 = -1##

Second Limit
##lim_(x->1) (x^n-1)/(x-1)##
We cannot use the quotient rule for limits, because the denominator has a limit of ##0##.
Because ##1## is a zero of the numerator, ##x-1## is a factor of the numerator. Either recall the factorization or use division to get:
##x^n-1 = (x-1)(x^(n-1)+x^(n-2)+ * * * +x^2+x+1)##
So,
##lim_(x->1) (x^n-1)/(x-1) =lim_(x->1) (x^(n-1)+x^(n-2)+ * * * +x^2+x+1)##
There are ##n## terms in the expression, each evaluating to ##1##, so, we get

## = underbrace(1+1+1+ * * * +1)_(n” terms”)##
## = n##