# what is the limit of((x^(p+1))-(p+1)*x+p)/((x-1)^2) when x approaching to 1, notice that p is in N set.

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Write((x^(p+1))-(p+1)*x+p)/((x-1)^2) as:

{x^(p+1)-x - p(x-1)}/(x-1)^2

={x(xP-1)-p(x-1)}/(x-1)^2

=(x-1){x(x^P-1 +x^p-2 +...x +1)-p}/(x-1)^2

= {x(x^P-1 +x^p-2 +...x +1)-p}/(x-1)

={(x^p +x^p-1 +x^p-2 +.....x) - p}/(x-1)

={(x^p -1)+(x^p-1 - 1)+(xp-2 -1)+...(x^2 -1)+(x-1)}/(x-1)....(1)

We know Lt a->1(a^n - 1)/(a-1) = Lt a->1(a^n-1 + a^n-2 +a^n-3+...a+1) = n-1. Applying this in (1) and taking limit as x->1, we get:

Lt x->1{(x^p -1)+(x^p-1 - 1)+(xp-2 -1)+...(x^2 -1)+(x-1)}/(x-1) = Lt x->0{(x^p-1)/x-1 + (x^p-1 -)/(x-1)+....(x^2-1)/(x-1)/(x-1)+(x-1)/(x-1)}

= p+p-1+p-2+p-3+....3+2+1

=pp+1)/2.

Therefore **Lt x->1 ((x^(p+1))-(p+1)*x+p)/((x-1)^2) = p(p+1)/2**.

without using hopital rule

I don't think so, because our teacher state that the limit of this functionm is: (p*(p+1))/2,

even if we use the hopital rule we get this result.

thanks anyway for the answer.