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:

• 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:

• $\lim_{x \to 0} \frac{\sqrt[m]{1 + ax} - \sqrt[n]{1 + bx}}{x}$ 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 :).

Let $$f(x)=x+x^2+\cdots+x^n-n$$ then by the definition of the derivative we have

$$\lim_{x \to 1} \frac{x + x^2 + \dots + x^n - n}{x - 1}= \lim_{x \to 1}\frac{f(x)-f(1)}{x - 1}=f'(1)\\[10pt] = \left[ \vphantom{\frac11} 1 + 2x + 3x^2 + \cdots + nx^{n-1} \right]_{x=1} = \frac{n(n + 1)}{2}$$

Well, you can use that \begin{align} x-1&=\left(x-1\right)\cdot 1,\\ x^2-1&=\left(x-1\right)\cdot\left(x+1\right),\\ x^3-1&=\left(x-1\right)\cdot(x^2+x+1),\\ &{}\ \ \vdots\\ x^n-1&=\left(x-1\right)\cdot(x^{n-1}+\cdots+1),\\ \end{align} Now sum the left hand sides and the right hand sides, divide by $x-1$ and consider the limit $x\rightarrow 1$.

The given limit is $$\lim_{x\rightarrow 1}\frac{\sum_{k=1}^nx^k-n}{x-1}\\ =\lim_{x\rightarrow 1}\frac{\sum_{k=1}^n(x^k-1)}{x-1}\\ =\sum_{k=1}^n \lim_{x\rightarrow 1} \frac{(x^k-1)}{x-1}$$

Now, $$\lim_{x\rightarrow 1} \frac{(x^k-1)}{x-1}\\ =\lim_{x\rightarrow 1} (\sum_{j=0}^{k-1}x^j)=k$$

Hence the given limit becomes $$\sum_{k=1}^n k=\frac{n(n+1)}{2}$$