Other methods for a limit I know that I can compute the limit
$$
\lim_{x\to1}\frac{Nx^{N+1}-(N+1)x^N+1}{(x-1)^2}=\frac{N(N+1)}{2}
$$
using L'Hospital's rule (not one but two times) but I am looking for other ways. Are there any of them?
p.s.: the limit follows from a shortcut used in order to find the value of $$
\sum_{k=1}^Nkx^k\Big|_{x=1}.
$$
 A: First shift the variable,
$$
\lim_{x\to1}\frac{nx^{n+1}-(n+1)x^n+1}{(x-1)^2}
=\lim_{x\to0}\frac{n(x+1)^{n+1}-(n+1)(x+1)^n+1}{x^2},
$$
then expand the numerator by means of the Binomial development.
The constant term is $n\displaystyle\binom{n+1}0-(n+1)\displaystyle\binom{n}0+1=0$.
The linear term is $n\displaystyle\binom{n+1}1x-(n+1)\displaystyle\binom{n}1x=0$.
The quadratic term is $n\displaystyle\binom{n+1}2x^2-(n+1)\displaystyle\binom{n}2x^2=\color{green}{\dfrac{n(n+1)}2}x^2$.
Higher order terms do not contribute.
A: Make the change: $t=x-1$. 
Then:
$$\lim_{x\to1}\frac{Nx^{N+1}-(N+1)x^N+1}{(x-1)^2}
=\\
\lim_{t\to 0}\frac{N\color{red}{(t+1)^{N+1}}-(N+1)\color{blue}{(t+1)^N}+1}{t^2}=\\
\lim_{t\to 0}\frac{N(\color{red}{O(t^3)+{N+1\choose 2}t^2+{N+1\choose 1}t+1})-(N+1)(\color{blue}{O(t^3)+{N\choose 2}t^2+{N\choose 1}t+1})+1}{t^2}=\\
\lim_{t\to 0}\frac{N(\color{red}{\frac{(N+1)N}{2}t^2+(N+1)t+1})-(N+1)(\color{blue}{\frac{N(N-1)}{2}t^2+Nt+1})+1}{t^2}=\\
\lim_{t\to 0}\frac{N\color{red}{\frac{(N+1)N}{2}t^2}-(N+1)\color{blue}{\frac{N(N-1)}{2}t^2}}{t^2}=\\
\frac{N(N+1)}{2}.$$
A: $$Nx^{N+1}-(N+1)x^N+1 = Nx^N(x-1)-(x^N-1)$$
$$ = Nx^N(x-1)-(x-1)(x^{N-1}+...x^2+x+1)$$
$$ = (x-1)\Big(\underbrace{Nx^N-(x^{N-1}+...x^2+x+1)}_{p(x)}\Big)$$
Now $$p(x) = (x^N-x^{N-1})+...+\color{red}{(x^N-x^2)}+\color{blue}{(x^N-x)}+\color{green}{(x^N-1)}$$
$$= x^{N-1}(x-1)+...+\color{red}{ x^2(x^{N-2}-1)}+\color{blue}{ x(x^{N-1}-1)}+\color{green}{(x^N-1)}$$
$$=(x-1)\Big(x^{N-1}+...+\color{red}{ x^2(x^{N-3}+...+x^2+x+1)}+\color{blue}{x(x^{N-2}+...+x^2+x+1)}+\color{green}{(x^{N-1}+...+x^2+x+1)}\Big)$$
So $$...=\lim_{x\to1}\Big(x^{N-1}+...+x^2(x^{N-3}+...+x^2+x+1)+x(x^{N-2}+...+x^2+x+1)+(x^{N-1}+...+x^2+x+1)\Big)$$
$$ 1+2+...+(N-2)+(N-1)+N=...$$
A: You can simplify your fraction as
\begin{align}
\frac{Nx^{N+1}-Nx^N+1-x^N}{(1-x)^2} &=\frac{-Nx^N(1-x)+(1-x)(1+x+\dots+x^{N-1})}{(1-x)^2} \\
& = \frac{-Nx^N+1+x+\dots+x^{N-1}}{1-x} \\
& = \frac{1+x+\dots+x^{N-1}+x^N-(N+1)x^N}{1-x}
\end{align}
For $x \to 1$ the numerator tends to $0$, so this is still a form $0/0$ and you can apply de l'Hôpital:
\begin{align}
\lim_{x \to 1} -(1+2x+3x^2+\dots+Nx^{N-1}-N(N+1)x^{N-1})& =-\sum_{k=1}^Nk+N(N+1) \\
& = N(N+1)/2.
\end{align}
