The value of lim _{x rightarrow 1} frac{x+x^2 | vedclass

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(C) We are given the limit $L = \lim _{x \rightarrow 1} \frac{x+x^2+\ldots+x^n-n}{x-1}$.Since the form is $\frac{0}{0}$ at $x=1$,we can rewrite the nu

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

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(C) We are given the limit $L = \lim _{x \rightarrow 1} \frac{x+x^2+\ldots+x^n-n}{x-1}$.Since the form is $\frac{0}{0}$ at $x=1$,we can rewrite the numerator as $(x-1) + (x^2-1) + \ldots + (x^n-1)$.Thus,$L = \lim _{x \rightarrow 1} \left( \frac{x-1}{x-1} + \frac{x^2-1}{x-1} + \ldots + \frac{x^n-1}{x-1} \right)$.Using the standard limit formula $\lim _{x \rightarrow a} \frac{x^n-a^n}{x-a} = na^{n-1}$,each term becomes $\lim _{x \rightarrow 1} \frac{x^k-1}{x-1} = k(1)^{k-1} = k$.Therefore,$L = 1 + 2 + 3 + \ldots + n$.The sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$.

The value of $\lim _{x \rightarrow 1} \frac{x+x^2+\ldots+x^n-n}{x-1}$ is

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