Re : Solve this sum of exponents under Limits

• We first substitute $x=1$

• We get a $\dfrac{0}{0}$ situation, so we can apply L'Hôpital's rule.

• We differentiate both numerator and denominator to get:

$\lim \limits_{x\rightarrow 1} \dfrac{x+x^2+x^3 \cdots x^{100} -100}{x-1} = \dfrac{\frac{d}{dx} (x+x^2+x^3 \cdots x^{100} -100)}{\frac{d}{dx} (x-1)}=\dfrac{1+2x+3x^2+4x^3 \cdots +100x^{99}}{1}$

• Now we can substitute $x=1$ and get the answer.

$\lim \limits_{x\rightarrow 1} \dfrac{x+x^2+x^3 \cdots x^{100} -100}{x-1} =\dfrac{100(101)}{2}=5050$

$\begin{aligned} L & = \lim_{x \to 1} \frac {x+x^2+x^3 + \cdots + x^{100}-100}{x-1} \\ & = \lim_{x \to 1} \frac {(x-1)+(x^2-1)+(x^3-1) + \cdots + (x^{100}-1)}{x-1} \\ & = \lim_{x to 1} \frac {(x-1)+(x-1)(x+1)+(x-1)(x^2+x+1)+\cdots + (x-1)(x^{99}+x^{98}+ \dots x + 1)}{x-1} \\ & = \lim_{x to 1} 1 + (x+1) + (x^2+x+1) + \cdots + (x^{99} + x^{98} + \cdots + x + 1) \\ & = 1 + 2 + 3 + \cdots + 100 = \frac {100(101)}2 = \boxed{5050} \end{aligned}$

The numerator can be written as

(x-1) + (^2 -1) + ( x^3 - 1) + (x^4 - 1).......(x^100 - 1)

This can be rewritten as

(x-1) + (x-1)(x+1) + (x-1)(x^2 + x + 1) + (x-1)(x^3+x^2+x+1).......(x-1)(x^99 + x^98..+1)

The expression when computed at x = 1 evaluates to ( x-1)(1 +2 + 3+ 4...100)/(x-1) under limits of x = 1

Under Limits that x = 1 , this computes to 100*101/2 = 5050