The value of `lim_(ntooo) [1/n+2]` is

Answer

C

Explanation

`L=underset(xto1)lim(nx^(n)(x-1)-(x^(n)-1))/((e^(x)-e)sinpix)`
`"Put "x=1+h" so that as "xto1,hto0." Therefore"`
`L=-underset(hto0)lim(h.n(1+h)^(n)-("("1+h")"^(n)-1))/(e(e^(h)-1)sinpih)`
`-underset(hto0)lim(n.h(1+""^(n)C_(1)h+""^(n)C_(2)h^(2)+""^(n)C_(3)h^(3)+...)-(1+""^(n)C_(1)h+""^(n)C_(2)h^(2)+""^(n)C_(3)h^(3)+...-1))/(pie(h^(2))((e^(h)-1)/(h))((sinpih)/(pih)))`
`=-(n^(2)-""^(n)C_(2))/(pie)=-[(2n^(2)-n(n-1))/(2pie)]`
`=-(n^(2)+n)/(2(pie))=-(n(n-1))/(2(pie))`
`" If "n=100," then "L=-((5050)/(pie)).`