If lim_{x to 1} frac{sin(3x^2 - 4x + 1) - x^2 | vedclass

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(C) Since the limit exists and is finite, the denominator must be $0$ at $x = 1$. Thus, $2(1)^3 - 7(1)^2 + a(1) + b = 0 \implies a + b = 5$.Using $L$'

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$x^2 - 5x - 24 = 0$

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(C) Since the limit exists and is finite, the denominator must be $0$ at $x = 1$. Thus, $2(1)^3 - 7(1)^2 + a(1) + b = 0 \implies a + b = 5$.Using $L$'$H$ôpital's rule, $\lim_{x 	o 1} \frac{\cos(3x^2 - 4x + 1)(6x - 4) - 2x}{6x^2 - 14x + a} = -2$.Substituting $x = 1$: $\frac{\cos(0)(2) - 2}{6 - 14 + a} = -2 \implies \frac{2 - 2}{a - 8} = -2$. This implies the numerator derivative is $0$, so we apply $L$'$H$ôpital's rule again.Derivative of numerator: $-\sin(3x^2 - 4x + 1)(6x - 4)^2 + \cos(3x^2 - 4x + 1)(6) - 2$.At $x = 1$: $-0 + 1(6) - 2 = 4$.Derivative of denominator: $12x - 14$.At $x = 1$: $12 - 14 = -2$.Limit $= \frac{4}{-2} = -2$. This confirms the limit value is independent of $a$ and $b$ as long as $a+b=5$ and the denominator is $0$ at $x=1$. However, for the limit to be defined, the denominator must have a factor $(x-1)$. Since $2x^3 - 7x^2 + ax + b = (x-1)(2x^2 - 5x + (a-5))$, for the limit to exist, $2x^2 - 5x + (a-5)$ must be divisible by $(x-1)$, so $2 - 5 + a - 5 = 0 \implies a = 8$. Then $b = 5 - 8 = -3$.Roots are $8$ and $-3$. The equation is $(x - 8)(x + 3) = x^2 - 5x - 24 = 0$.

If $\lim_{x \to 1} \frac{\sin(3x^2 - 4x + 1) - x^2 + 1}{2x^3 - 7x^2 + ax + b} = -2$, then the quadratic equation having roots $a$ and $b$ is

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