##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##

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