Key Concepts and Formulas

• Limit of a function: The value a function approaches as the input approaches some value.
• Standard Limit $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$: This is a fundamental limit used when dealing with trigonometric functions in limits.
• L'Hôpital's Rule: If a limit results in an indeterminate form (like $\frac{0}{0}$ or $\frac{\infty}{\infty}$), we can differentiate the numerator and denominator separately and then evaluate the limit.
• Polynomial roots: If a polynomial $P(x)$ has $x=c$ as a root, then $P(c)=0$. If $x=c$ is a repeated root of multiplicity $m$, then $P(c) = P'(c) = \dots = P^{(m-1)}(c) = 0$.

Step-by-Step Solution

Let the given limit be $L$. $L = \mathop {\lim }\limits_{x \to 1} {{\sin (3{x^2} - 4x + 1) - {x^2} + 1} \over {2{x^3} - 7{x^2} + ax + b}} = - 2$

Step 1: Analyze the numerator as $x \to 1$. As $x \to 1$, the argument of the sine function, $3x^2 - 4x + 1$, approaches $3(1)^2 - 4(1) + 1 = 3 - 4 + 1 = 0$. Also, as $x \to 1$, the term $-x^2 + 1$ approaches $-(1)^2 + 1 = 0$. Thus, the numerator approaches $\sin(0) - 0 = 0$.

Step 2: Analyze the denominator as $x \to 1$. For the limit to be a finite non-zero value (like -2), the denominator must also approach 0 as $x \to 1$. Otherwise, if the denominator approached a non-zero value, the limit would be 0. So, let $D(x) = 2x^3 - 7x^2 + ax + b$. We must have $D(1) = 0$. $2(1)^3 - 7(1)^2 + a(1) + b = 0$ $2 - 7 + a + b = 0$ $a + b - 5 = 0 \quad \Rightarrow \quad a + b = 5$

Step 3: Apply the standard limit formula and simplify the expression. We can rewrite the numerator using the standard limit $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$. Let $\theta = 3x^2 - 4x + 1$. As $x \to 1$, $\theta \to 0$. $L = \mathop {\lim }\limits_{x \to 1} \frac{\frac{\sin (3{x^2} - 4x + 1)}{3{x^2} - 4x + 1} \cdot (3{x^2} - 4x + 1) - ({x^2} - 1)}{2{x^3} - 7{x^2} + ax + b} = - 2$ Since $\mathop {\lim }\limits_{x \to 1} \frac{\sin (3{x^2} - 4x + 1)}{3{x^2} - 4x + 1} = 1$, the limit becomes: $L = \mathop {\lim }\limits_{x \to 1} \frac{1 \cdot (3{x^2} - 4x + 1) - ({x^2} - 1)}{2{x^3} - 7{x^2} + ax + b} = - 2$ Simplify the numerator: $(3{x^2} - 4x + 1) - ({x^2} - 1) = 3x^2 - 4x + 1 - x^2 + 1 = 2x^2 - 4x + 2$ Factor the numerator: $2x^2 - 4x + 2 = 2(x^2 - 2x + 1) = 2(x-1)^2$ So, the limit expression is: $L = \mathop {\lim }\limits_{x \to 1} \frac{2(x-1)^2}{2{x^3} - 7{x^2} + ax + b} = - 2$

Step 4: Analyze the denominator further using the fact that the limit is finite. We know that the numerator has a factor of $(x-1)^2$. For the limit to be finite and non-zero, the denominator must have at least a factor of $(x-1)^2$. This means $x=1$ must be a repeated root of the denominator polynomial $D(x) = 2x^3 - 7x^2 + ax + b$. If $x=1$ is a repeated root, then $D(1) = 0$ and $D'(1) = 0$. We already used $D(1)=0$ to get $a+b=5$.

Now, let's find the derivative of $D(x)$: $D'(x) = \frac{d}{dx}(2{x^3} - 7{x^2} + ax + b) = 6x^2 - 14x + a$ Set $D'(1) = 0$: $6(1)^2 - 14(1) + a = 0$ $6 - 14 + a = 0$ $-8 + a = 0$ $a = 8$

Step 5: Find the value of b. Using the equation from Step 2, $a + b = 5$, substitute $a=8$: $8 + b = 5$ $b = 5 - 8$ $b = -3$

Step 6: Verify the limit with the found values of a and b. With $a=8$ and $b=-3$, the denominator is $D(x) = 2x^3 - 7x^2 + 8x - 3$. We need to check if $(x-1)^2$ is a factor of $D(x)$. $D(1) = 2(1)^3 - 7(1)^2 + 8(1) - 3 = 2 - 7 + 8 - 3 = 0$. $D'(x) = 6x^2 - 14x + 8$. $D'(1) = 6(1)^2 - 14(1) + 8 = 6 - 14 + 8 = 0$. Since $D(1)=0$ and $D'(1)=0$, $x=1$ is at least a double root. We can perform polynomial division or synthetic division to factor $D(x)$. Since $(x-1)^2 = x^2 - 2x + 1$ is a factor, we can divide $2x^3 - 7x^2 + 8x - 3$ by $x^2 - 2x + 1$.

Alternatively, we can use L'Hôpital's Rule on the limit expression from Step 3: $L = \mathop {\lim }\limits_{x \to 1} \frac{2(x-1)^2}{2{x^3} - 7{x^2} + 8x - 3}$ This is of the form $\frac{0}{0}$. Apply L'Hôpital's Rule: $L = \mathop {\lim }\limits_{x \to 1} \frac{\frac{d}{dx}(2(x-1)^2)}{\frac{d}{dx}(2{x^3} - 7{x^2} + 8x - 3)} = \mathop {\lim }\limits_{x \to 1} \frac{4(x-1)}{6x^2 - 14x + 8}$ This is still of the form $\frac{0}{0}$. Apply L'Hôpital's Rule again: $L = \mathop {\lim }\limits_{x \to 1} \frac{\frac{d}{dx}(4(x-1))}{\frac{d}{dx}(6x^2 - 14x + 8)} = \mathop {\lim }\limits_{x \to 1} \frac{4}{12x - 14}$ Now substitute $x=1$: $L = \frac{4}{12(1) - 14} = \frac{4}{12 - 14} = \frac{4}{-2} = -2$ This matches the given limit value, confirming our values of $a=8$ and $b=-3$.

Step 7: Calculate the value of (a - b). $a - b = 8 - (-3) = 8 + 3 = 11$

Common Mistakes & Tips

• Forgetting the condition for finite limits: If the limit is finite and non-zero, and the numerator approaches zero, the denominator must also approach zero. This is crucial for solving for the unknown constants.
• Incorrect application of standard limits: Ensure the argument of the sine function matches the denominator when using $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.
• Mistakes in differentiation or polynomial factorization: Careful calculation is needed when applying L'Hôpital's Rule or factoring polynomials. Checking the roots and derivatives of the denominator polynomial is a robust method.

Summary

The problem involves evaluating a limit with unknown constants in the denominator. We first established that for the limit to be finite, the denominator must be zero at $x=1$. By applying the standard limit $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ and simplifying the numerator, we found that the limit reduces to $\mathop {\lim }\limits_{x \to 1} \frac{2(x-1)^2}{D(x)} = -2$. This implies that $x=1$ must be a repeated root of the denominator polynomial $D(x)$. This condition yields two equations by setting $D(1)=0$ and $D'(1)=0$, which allowed us to solve for the constants $a$ and $b$. Finally, we calculated $a-b$.

The final answer is $\boxed{11}$.