Question

Find the following limit.
$\underset{x\to 1}{lim}\frac{x+x^2+...+x^n-n}{x-1}.$

Answer 1

$\underset{x\to 1}{lim}\frac{x+x^2+x^3....+x^n-n}{x-1}$

here $\frac{0}{0}$ form occurs so from L Hospital Rule

$\underset{x\to 1}{lim}\frac{\frac{d}{dx}(x+x^2+x^3+.....+x^n-n)}{\frac{d}{dx}(x-1)}$

$\underset{x\to 1}{lim}\frac{1+2x+3x+.....+n{x}^{n-1}-0}{1-0}$

$=\frac{1+2+3+....+n}{1}[1+2+3+.....+n=\sum n=\frac{n(n+1)}{2}]$

$\frac{n(n+1)}{2}$.