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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 $$ \sum_{k=1}^Nkx^k\Big|_{x=1}. $$

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    $\begingroup$ There is no factor $x$ in the first limit, right? $\endgroup$ – Gibbs Jul 3 '19 at 9:42
  • $\begingroup$ @Gibbs you were right! thanks $\endgroup$ – yngabl Jul 3 '19 at 9:45
  • $\begingroup$ Instead of applying de L'Hôpital twice you can simplify the numerator and then apply de l'Hôpital only once. If that answers your question, I can post my solution as an answer. $\endgroup$ – Gibbs Jul 3 '19 at 9:50
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    $\begingroup$ Isn't the limit for $\sum_{k=1}^N k$ ? $\endgroup$ – Yves Daoust Jul 3 '19 at 10:28
  • $\begingroup$ @YvesDaoust of course $\endgroup$ – yngabl Jul 3 '19 at 10:45
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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.

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Make the change: $t=x-1$.

Then: $$\lim_{x\to1}\frac{Nx^{N+1}-(N+1)x^N+1}{(x-1)^2} =\\ \lim_{t\to 0}\frac{N\color{red}{(t+1)^{N+1}}-(N+1)\color{blue}{(t+1)^N}+1}{t^2}=\\ \lim_{t\to 0}\frac{N(\color{red}{O(t^3)+{N+1\choose 2}t^2+{N+1\choose 1}t+1})-(N+1)(\color{blue}{O(t^3)+{N\choose 2}t^2+{N\choose 1}t+1})+1}{t^2}=\\ \lim_{t\to 0}\frac{N(\color{red}{\frac{(N+1)N}{2}t^2+(N+1)t+1})-(N+1)(\color{blue}{\frac{N(N-1)}{2}t^2+Nt+1})+1}{t^2}=\\ \lim_{t\to 0}\frac{N\color{red}{\frac{(N+1)N}{2}t^2}-(N+1)\color{blue}{\frac{N(N-1)}{2}t^2}}{t^2}=\\ \frac{N(N+1)}{2}.$$

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$$Nx^{N+1}-(N+1)x^N+1 = Nx^N(x-1)-(x^N-1)$$ $$ = Nx^N(x-1)-(x-1)(x^{N-1}+...x^2+x+1)$$ $$ = (x-1)\Big(\underbrace{Nx^N-(x^{N-1}+...x^2+x+1)}_{p(x)}\Big)$$

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 $$...=\lim_{x\to1}\Big(x^{N-1}+...+x^2(x^{N-3}+...+x^2+x+1)+x(x^{N-2}+...+x^2+x+1)+(x^{N-1}+...+x^2+x+1)\Big)$$ $$ 1+2+...+(N-2)+(N-1)+N=...$$

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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}

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