Question

$\underset{x\to 1}{lim}\frac{x^{n+1}-(n+1)x+n}{(x-1{)}^2}$.

Answer 1

${lim}_{x\to 1}\frac{x^{n+1}-(n+1)x+n}{{(x-1)}^2}=\frac{1-(n+1)+n}{{(1-1)}^2}$

$=\frac{0}{0}$ form

Using L-Hospital's rule

$={lim}_{x\to 1}\frac{(n+1)x^{n+1}-(n+1)+0}{2(x-1)}$

$={lim}_{x\to 1}\frac{n+1}{2}\times \frac{x^n-1}{x-1}$

$=\frac{n+1}{2}\times n.1^{n-1}=\frac{n(n+1)}{2}$