How to show $\lim_{x \to 1} \frac{x + x^2 + \dots + x^n - n}{x - 1} = \frac{n(n + 1)}{2}$? I am able to evaluate the limit $$\lim_{x \to 1} \frac{x + x^2 + \dots + x^n - n}{x - 1} = \frac{n(n + 1)}{2}$$ for a given $n$ using l'Hôspital's (Bernoulli's) rule.
The problem is I don't quite like the solution, as it depends on such a heavy weaponry. A limit this simple, should easily be evaluable using some clever idea. Here is a list of what I tried:


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*Substitute $y = x - 1$. This leads nowhere, I think.

*Find the Taylor polynomial. Makes no sense, it is a polynomial.

*Divide by major term. Dividing by $x$ got me nowhere.

*Find the value $f(x)$ at $x = 1$ directly. I cannot as the function is not defined at $x = 1$.

*Simplify the expression. I do not see how I could.

*Using l'Hôspital's (Bernoulli's) rule. Works, but I do not quite like it.


If somebody sees a simple way, please do let me know.

Added later: The approach proposed by Sami Ben Romdhane is universal as asmeurer pointed out. Examples of another limits that can be easily solved this way:


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*$\lim_{x \to 0} \frac{\sqrt[m]{1 + ax} - \sqrt[n]{1 + bx}}{x}$ where $m, n \in \mathbb{N}$ and $a, b \in \mathbb{R}$ are given, or

*$\lim_{x \to 0} \frac{\arctan(1 + x) - \arctan(1 - x)}{x}$.


It sems that all limits in the form $\lim_{x \to a} \frac{f(x)}{x - a}$ where $a \in \mathbb{R}$, $f(a) = 0$ and for which $\exists f'(a)$, can be evaluated this way, which is as fast as finding $f'$ and calculating $f'(a)$.
This adds a very useful tool into my calculus toolbox: Some limits can be evaluated easily using derivatives if one looks for $f(a) = 0$, without the l'Hôspital's rule. I have not seen this in widespread use; I propose we call this Sami's rule :).

Hints:
$$x+x^2+\ldots+x^n-n=(x-1)+(x^2-1)+\cdots +(x^n-1)=(x-1)\left(1+(x+1)+\cdots\right)$$
Also
$$1+(x+1)+(x^2+x+1)+\cdots+(x^{n-1}+\cdots+x+1)\xrightarrow [x\to 1]{}1+2+\cdots+n=\frac{n(n+1)}2$$
