Solve $\lim \:_{x\to \:1}\frac{1+x+x^2+...\:+x^n-\left(n+1\right)}{x-1}$ How do I solve limits such as these? The $...$ always make it seem hard to me. From what I can understand from them, they both are $0/0$ limits, and I should be looking to write the numerator in such a way that the denominator should simplify it somehow. For the first I tried writing each element from the numerator as $(a-1)$ where $a = 1, x, x^2 ... x^n$ so that I can simplify the denominator eventually, but I didn't get much out of that. I'm guessing these two are solved in similar ways, hence why I posted them both. Any clues/hints I can get would be appreciated.
$$\lim \:_{x\to \:1}\frac{1+x+x^2+...\:+x^n-\left(n+1\right)}{x-1}$$
$$\lim _{x\to 1}\frac{x+x^2+...\:x^n-n}{x+x^2+...\:+x^m-m}$$
$n,\:m\:\in \mathbb{R}$

Note that:
$\frac{1+x+x^2+...\:+x^n-\left(n+1\right)}{x-1}= \frac{(x-1)+(x^2-1)+\ldots +(x^n-1)}{x-1}=\frac{(x-1)[1+(x+1)+(x^2+x+1)\ldots +(x^{n-1}+x^{n-2}+\ldots+x+1)]}{x-1}$,
after simplification we find:
$n+(n-1)x+(n-2)x^2+\ldots+x^{n-1}$.
So,
$\lim \:_{x\to \:1}\frac{1+x+x^2+...\:+x^n-\left(n+1\right)}{x-1}=n+(n-1)+(n-2)+\ldots+2+1=\frac{n(n+1)}{2}$.
For the second one, we use the following method:
$\lim _{x\to 1}\frac{x+x^2+...\:x^n-n}{x+x^2+...\:+x^m-m}=\lim _{x\to 1}\frac{x+x^2+...\:x^n-n}{x-1}\times\frac{x-1}{x+x^2+...\:+x^m-m}=\frac{n(n+1)}{2}\times\frac{2}{m(m+1)}=\frac{n(n+1)}{m(m+1)}$
