Key Concepts and Formulas

• Limit of a Rational Function: If the limit of a rational function of the form $\mathop {\lim }\limits_{x \to c} \frac{P(x)}{Q(x)}$ results in an indeterminate form like $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then further evaluation is required.
• Condition for Finite Limit: For the limit $\mathop {\lim }\limits_{x \to c} \frac{P(x)}{Q(x)}$ to be finite when $Q(c) = 0$, it is necessary that $P(c) = 0$. This ensures the indeterminate form $\frac{0}{0}$, allowing for methods like factorization or L'Hôpital's Rule.
• L'Hôpital's Rule: If $\mathop {\lim }\limits_{x \to c} \frac{f(x)}{g(x)}$ results in an indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\mathop {\lim }\limits_{x \to c} \frac{f(x)}{g(x)} = \mathop {\lim }\limits_{x \to c} \frac{f'(x)}{g'(x)}$, provided the latter limit exists.

Step-by-Step Solution

Step 1: Analyze the Limit and Identify Indeterminate Form We are given the limit: $\mathop {\lim }\limits_{x \to 1} \frac{x^2 - ax + b}{x - 1} = 5$ As $x \to 1$, the denominator $(x - 1) \to 0$. For the limit to be a finite value (5 in this case), the numerator must also approach 0 as $x \to 1$. If the numerator approached a non-zero value, the limit would be infinite. This is a crucial condition for a finite limit when the denominator is zero.

Step 2: Apply the Condition for a Finite Limit Since the denominator approaches 0 as $x \to 1$, the numerator must also be 0 at $x = 1$ for the limit to be finite. Substitute $x = 1$ into the numerator: $(1)^2 - a(1) + b = 0$ $1 - a + b = 0$ Rearranging this equation, we get our first relationship between $a$ and $b$: $a - b = 1 \quad \cdots (1)$

Step 3: Apply L'Hôpital's Rule Since the limit resulted in the indeterminate form $\frac{0}{0}$ at $x=1$, we can apply L'Hôpital's Rule. This involves taking the derivative of the numerator and the derivative of the denominator with respect to $x$. The derivative of the numerator, $x^2 - ax + b$, is $2x - a$. The derivative of the denominator, $x - 1$, is $1$. Now, we can rewrite the limit using L'Hôpital's Rule: $\mathop {\lim }\limits_{x \to 1} \frac{2x - a}{1} = 5$

Step 4: Evaluate the Limit After Applying L'Hôpital's Rule Now, substitute $x = 1$ into the simplified expression: $\frac{2(1) - a}{1} = 5$ $2 - a = 5$ Solving for $a$: $a = 2 - 5$ $a = -3$

Step 5: Find the Value of b We have found the value of $a$. Now, we can substitute this value into Equation (1) to find the value of $b$. From Equation (1): $a - b = 1$ Substitute $a = -3$: $-3 - b = 1$ Solving for $b$: $-b = 1 + 3$ $-b = 4$ $b = -4$

Step 6: Calculate a + b The question asks for the value of $a + b$. We have $a = -3$ and $b = -4$. $a + b = (-3) + (-4)$ $a + b = -7$

Common Mistakes & Tips

• Forgetting the $\frac{0}{0}$ Condition: A common mistake is to directly apply L'Hôpital's Rule without first verifying that the limit results in an indeterminate form. If the denominator is 0 and the numerator is not 0, the limit is infinite, and L'Hôpital's Rule is not applicable.
• Algebraic Errors: Mistakes in solving the linear equations for $a$ and $b$ can lead to an incorrect final answer. Double-check your calculations, especially when dealing with negative signs.
• Alternative Method (Factorization): If you recognize that for the limit to be finite, $x-1$ must be a factor of the numerator $x^2 - ax + b$, you can factor the numerator. Since $x=1$ is a root, $(x-1)$ is a factor. Let the other factor be $(x-k)$. Then $x^2 - ax + b = (x-1)(x-k) = x^2 - (1+k)x + k$. Comparing coefficients, $a = 1+k$ and $b=k$. Substituting $x=1$ into the numerator yields $1-a+b=0$, so $a-b=1$. Substituting $b=k$ and $a=1+k$ gives $(1+k)-k=1$, which is always true. The limit becomes $\mathop {\lim }\limits_{x \to 1} \frac{(x-1)(x-k)}{x-1} = \mathop {\lim }\limits_{x \to 1} (x-k) = 1-k$. Since this limit is 5, $1-k=5$, so $k=-4$. Then $b=k=-4$, and $a=1+k = 1+(-4) = -3$. This gives $a=-3$ and $b=-4$, leading to $a+b=-7$.

Summary

The problem involves evaluating a limit of a rational function that results in an indeterminate form. We first established that for the limit to be finite, the numerator must be zero when $x=1$, leading to an equation relating $a$ and $b$. Subsequently, we applied L'Hôpital's Rule to simplify the limit expression and found the value of $a$. Finally, we used the equation from the first step to determine the value of $b$, and then calculated the sum $a+b$.

The final answer is \boxed{-7}.