Find the Limit

$\begin{aligned} L & = \lim_{x \to 1} \frac {\sum_{k=1}^{100}-1}{x-1} \\ & = \lim_{x \to 1} \frac {x+x^2+x^3 + \cdots + x^{100} - 100}{x-1} & \small \blue{\text{A 0/0 case, L'Hôpital's rule applies.}} \\ & = \lim_{x \to 1} \frac {1+2x+3x^2 + \cdots + 100x^{99}}{1} & \small \blue{\text{Differentiate up and down w.r.t. }x} \\ & = \frac {100(101)}2 = \boxed{5050} \end{aligned}$

Reference: L'Hôpital's rule

The given function can be simplified to $\dfrac{x^{101}-101x+100}{(x-1)^2}$ . Since in the given limit the function assumes the form $\dfrac{0}{0}$ , we apply L'Hospital's rule twice to obtain the limit as $\dfrac{101\times 100}{2}=\boxed {5050}$ .

Here's a non-L'Hôpital approach. If we rearrange the sum slightly, we can use direct algebraic division:

$\begin{aligned} \lim_{x \to 1} \frac{\sum_{k=1}^{100} x^k-100}{x-1} &=\lim_{x \to 1} \frac{\sum_{k=1}^{100} \left(x^k-1\right)}{x-1} \\ &=\lim_{x \to 1} \left(1+(x+1)+\left(x^2+x+1\right)+\cdots+\left(x^{99}+\cdots+x+1\right) \right) \\ &=1+2+3+\cdots+100 \\ &=\boxed{5050} \end{aligned}$