lim _{x rightarrow 1} left( frac{x+x^2+x^3+ld | vedclass

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(A) Let $L = \lim _{x \rightarrow 1} \left( \frac{x+x^2+x^3+\ldots+x^n-n}{x-1} \right)$.We can rewrite the numerator as: $(x-1) + (x^2-1) + (x^3-1) +

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$\frac{n(n+1)}{2}$

Vedclass · Answer

(A) Let $L = \lim _{x \rightarrow 1} \left( \frac{x+x^2+x^3+\ldots+x^n-n}{x-1} \right)$.We can rewrite the numerator as: $(x-1) + (x^2-1) + (x^3-1) + \ldots + (x^n-1)$.So,$L = \lim _{x \rightarrow 1} \left( \frac{(x-1) + (x^2-1) + (x^3-1) + \ldots + (x^n-1)}{x-1} \right)$.$L = \lim _{x}$ ${\rightarrow 1} \left( \frac{x-1}{x-1} + \frac{x^2-1}{x-1} + \frac{x^3-1}{x-1} + \ldots + \frac{x^n-1}{x-1} \right)$.Using the standard limit $\lim _{x \rightarrow a} \frac{x^n-a^n}{x-a} = na^{n-1}$,we get:$L = 1 + 2(1)^{2-1} + 3(1)^{3-1} + \ldots + n(1)^{n-1}$.$L = 1 + 2 + 3 + \ldots + n$.$L = \frac{n(n+1)}{2}$.

$\lim _{x \rightarrow 1} \left( \frac{x+x^2+x^3+\ldots+x^n-n}{x-1} \right) = $

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