Answer
$L(n)=\dfrac{n-1}{2}$, $n$ integer
Work Step by Step
We have to determine the limit:
$L(n)=\displaystyle\lim_{x\rightarrow 1} \left(\dfrac{n}{1-x^n}-\dfrac{1}{1-x}\right)$
Determine $L(n)$ for $n=1,2,3$:
$L(1)=\displaystyle\lim_{x\rightarrow 1} \left(\dfrac{1}{1-x^1}-\dfrac{1}{1-x}\right)=0$
$L(2)=\displaystyle\lim_{x\rightarrow 1} \left(\dfrac{2}{1-x^2}-\dfrac{1}{1-x}\right)$
$=\displaystyle\lim_{x\rightarrow 1} \left(\dfrac{2-1-x}{1-x^2}\right)$
$=\displaystyle\lim_{x\rightarrow 1} \left(\dfrac{1-x}{(1-x)(1+x)}\right)$
$=\displaystyle\lim_{x\rightarrow 1} \left(\dfrac{1}{1+x}\right)$
$=\dfrac{1}{2}$
$L(3)=\displaystyle\lim_{x\rightarrow 1} \left(\dfrac{3}{1-x^3}-\dfrac{1}{1-x}\right)$
$=\displaystyle\lim_{x\rightarrow 1} \left(\dfrac{3}{(1-x)(1+x+x^2)}-\dfrac{1}{1-x}\right)$
$=\displaystyle\lim_{x\rightarrow 1} \left(\dfrac{3-1-x-x^2}{(1-x)(1+x)}\right)$
$=\displaystyle\lim_{x\rightarrow 1} \left(\dfrac{-x^2-x+2}{(1-x)(1+x+x^2)}\right)$
$=\displaystyle\lim_{x\rightarrow 1} \left(\dfrac{x+2}{1+x+x^2}\right)$
$=1$
$L(n)=\displaystyle\lim_{x\rightarrow 1} \left(\dfrac{n}{1-x^n}-\dfrac{1}{1-x}\right)$
$=\displaystyle\lim_{x\rightarrow 1} \left(\dfrac{n}{(1-x)(1+x+x^2+...+x^{n-1})}-\dfrac{1}{1-x}\right)$
$=\displaystyle\lim_{x\rightarrow 1} \left(\dfrac{n-1-x-x^2-...-x^{n-1}}{(1-x)(1+x+x^2+...+x^{n-1})}\right)$
$=\displaystyle\lim_{x\rightarrow 1} \left(\dfrac{(1-x)+(1-x^2)+...+(1-x^{n-1})}{(1-x)(1+x+x^2+...+x^{n-1})}\right)$
$=\displaystyle\lim_{x\rightarrow 1} \left(\dfrac{(1-x)(1+(1+x)+...+(1+x+...+x^{n-2})}{(1-x)(1+x+x^2+...+x^{n-1})}\right)$
$=\displaystyle\lim_{x\rightarrow 1} \left(\dfrac{1+(1+x)+...+(1+x+...+x^{n-2})}{1+x+x^2+...+x^{n-1}}\right)$
$=\dfrac{1+2+3+...+(n-1)}{1+1+...+1}$
$=\dfrac{\dfrac{(n-1)n}{2}}{n}$
$=\dfrac{n-1}{2}$
