If lim_{x to 1} frac{x^3 + ax^2 + bx + c}{x^2 | vedclass

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(D) For the limit to exist and be finite, the numerator must be $0$ at $x = 1$. Thus, $1 + a + b + c = 0$, which implies $b = -1 - a - c$.Substituting

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$-2$

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(D) For the limit to exist and be finite, the numerator must be $0$ at $x = 1$. Thus, $1 + a + b + c = 0$, which implies $b = -1 - a - c$.Substituting $b$ into the numerator: $x^3 + ax^2 - (1 + a + c)x + c = (x^3 - x) + a(x^2 - x) - c(x - 1) = (x - 1)(x^2 + x + ax - c)$.The limit becomes $\lim_{x 	o 1} \frac{(x - 1)(x^2 + (1 + a)x - c)}{(x - 1)^2} = \lim_{x 	o 1} \frac{x^2 + (1 + a)x - c}{x - 1}$.For this limit to be finite, the numerator must be $0$ at $x = 1$: $1 + (1 + a) - c = 0 \implies 2 + a - c = 0 \implies a - c = -2$.

If $\lim_{x \to 1} \frac{x^3 + ax^2 + bx + c}{x^2 - 2x + 1} = 2026$, then the value of $a - c$ is...

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