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Find the limit of the given function

Find the value of the following limit.

    \[ \lim_{x \to 1} \frac{ \sum_{k=1}^n x^k - n }{x-1}. \]


First, we use the algebraic identity,

    \[ \sum_{k=1}^n x^k = \frac{1-x^{n+1}}{1-x} - 1\]

to obtain

    \begin{align*}   \lim_{x \to 1} \frac{ \sum_{k=1}^n x^k - n }{x-1} &= \lim_{x \to 1} \frac{ \left( \frac{1-x^{n+1}}{1-x} \right) - 1 - n}{x-1} \\[9pt]  &= \lim_{x \to 1} \frac{ 1 - x^{n+1} - (1+n)(1-x)}{(1-x)(x-1)} \\[9pt]  &= \lim_{x \to 1} \frac{x^{n+1} - 1 + (1+n)(1-x)}{(x-1)^2}. \end{align*}

Then, we apply L’Hopital’s rule twice,

    \begin{align*}  \lim_{x \to 1} \frac{\sum_{k=1}^ x^k - n}{x-1} &= \lim_{x \to 1} \frac{x^{n+1} - 1 + (1+n)(1-x)}{(x-1)^2} \\[9pt]  &= \lim_{x \to 1} \frac{(n+1)x^n - (1+n)}{2(x-1)} \\[9pt]  &= \lim_{x \to 1} \frac{n(n+1)x^{n-1}}{2} \\[9pt]  &= \frac{n(n+1)}{2}. \end{align*}

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