Key Concepts and Formulas

• Limits at Infinity: When evaluating limits as $x \to \infty$, we often look at the highest power of $x$ in the numerator and denominator.
• Indeterminate Forms: The form $\infty - \infty$ is an indeterminate form that requires algebraic manipulation (like rationalization) to resolve.
• Rationalization: Multiplying the numerator and denominator by the conjugate is a common technique to handle square roots in limits.
• Standard Limit Results: Knowledge of limits like $\lim_{x \to \infty} \frac{1}{x^n} = 0$ for $n > 0$ is crucial.

Step-by-Step Solution

We are given the limit: $\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} - x + 1} - ax} \right) = b$

Step 1: Identify the Indeterminate Form As $x \to \infty$, $\sqrt{x^2 - x + 1}$ behaves like $\sqrt{x^2} = x$. Therefore, the expression inside the limit becomes $\infty - ax$. For the limit to be a finite value $b$, this must be an indeterminate form of the type $\infty - \infty$. This implies that the dominant terms must cancel out. Specifically, if $a > 1$, the limit would be $-\infty$, and if $a < 1$, the limit would be $+\infty$. Thus, for a finite limit $b$, we must have $a=1$.

Step 2: Rationalize the Expression To resolve the $\infty - \infty$ indeterminate form, we multiply the expression by its conjugate divided by itself. The conjugate of $\sqrt{x^2 - x + 1} - ax$ is $\sqrt{x^2 - x + 1} + ax$. $\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} - x + 1} - ax} \right) \times \frac{\sqrt{x^2 - x + 1} + ax}{\sqrt{x^2 - x + 1} + ax} = b$ $\mathop {\lim }\limits_{x \to \infty } \frac{(\sqrt{x^2 - x + 1})^2 - (ax)^2}{\sqrt{x^2 - x + 1} + ax} = b$ $\mathop {\lim }\limits_{x \to \infty } \frac{(x^2 - x + 1) - a^2x^2}{\sqrt{x^2 - x + 1} + ax} = b$ $\mathop {\lim }\limits_{x \to \infty } \frac{(1 - a^2)x^2 - x + 1}{\sqrt{x^2 - x + 1} + ax} = b$

Step 3: Analyze the Dominant Terms in the Numerator and Denominator To evaluate the limit as $x \to \infty$, we divide both the numerator and the denominator by the highest power of $x$ in the denominator. In the denominator, $\sqrt{x^2 - x + 1}$ behaves like $\sqrt{x^2} = x$, and $ax$ is also of order $x$. So, we divide by $x$. First, let's factor out $x^2$ from the term $(1 - a^2)x^2 - x + 1$ in the numerator: $\mathop {\lim }\limits_{x \to \infty } \frac{x^2((1 - a^2) - \frac{1}{x} + \frac{1}{x^2})}{\sqrt{x^2(1 - \frac{1}{x} + \frac{1}{x^2})} + ax} = b$ $\mathop {\lim }\limits_{x \to \infty } \frac{x^2((1 - a^2) - \frac{1}{x} + \frac{1}{x^2})}{x\sqrt{1 - \frac{1}{x} + \frac{1}{x^2}} + ax} = b$ Now, factor out $x$ from the denominator: $\mathop {\lim }\limits_{x \to \infty } \frac{x^2((1 - a^2) - \frac{1}{x} + \frac{1}{x^2})}{x(\sqrt{1 - \frac{1}{x} + \frac{1}{x^2}} + a)} = b$ $\mathop {\lim }\limits_{x \to \infty } \frac{x((1 - a^2) - \frac{1}{x} + \frac{1}{x^2})}{\sqrt{1 - \frac{1}{x} + \frac{1}{x^2}} + a} = b$

Step 4: Determine the Value of 'a' For the limit to be finite ($b$), the coefficient of the highest power of $x$ in the numerator must be zero. In the expression above, the numerator has a factor of $x$ multiplied by $(1-a^2)$. If $(1-a^2) \neq 0$, the numerator would grow like $x$, making the overall limit $\infty$ or $-\infty$. Therefore, we must have: $1 - a^2 = 0$ $a^2 = 1$ $a = \pm 1$ From Step 1, we deduced that $a=1$ is necessary for a finite limit. Let's confirm this by substituting $a=1$ back into the limit expression.

Step 5: Evaluate the Limit with the Determined Value of 'a' Substitute $a=1$ into the limit expression from Step 3: $\mathop {\lim }\limits_{x \to \infty } \frac{x((1 - 1^2) - \frac{1}{x} + \frac{1}{x^2})}{\sqrt{1 - \frac{1}{x} + \frac{1}{x^2}} + 1} = b$ $\mathop {\lim }\limits_{x \to \infty } \frac{x(0 - \frac{1}{x} + \frac{1}{x^2})}{\sqrt{1 - \frac{1}{x} + \frac{1}{x^2}} + 1} = b$ $\mathop {\lim }\limits_{x \to \infty } \frac{-\frac{1}{x} + \frac{1}{x^2}}{\sqrt{1 - \frac{1}{x} + \frac{1}{x^2}} + 1} = b$

Step 6: Calculate the Value of 'b' Now, we can evaluate the limit by substituting $\frac{1}{x} \to 0$ and $\frac{1}{x^2} \to 0$ as $x \to \infty$: $b = \frac{0 + 0}{\sqrt{1 - 0 + 0} + 1}$ $b = \frac{0}{\sqrt{1} + 1}$ $b = \frac{0}{1 + 1}$ $b = \frac{0}{2}$ $b = 0$ This result for $b$ is not matching the provided options. Let's re-examine Step 3 and Step 5.

Correction/Re-evaluation of Step 3 and Step 5:

Let's return to the expression after rationalization: $\mathop {\lim }\limits_{x \to \infty } \frac{(1 - a^2)x^2 - x + 1}{\sqrt{x^2 - x + 1} + ax} = b$ Divide numerator and denominator by $x$: $\mathop {\lim }\limits_{x \to \infty } \frac{\frac{(1 - a^2)x^2 - x + 1}{x}}{\frac{\sqrt{x^2 - x + 1} + ax}{x}} = b$ $\mathop {\lim }\limits_{x \to \infty } \frac{(1 - a^2)x - 1 + \frac{1}{x}}{\sqrt{\frac{x^2 - x + 1}{x^2}} + a} = b$ $\mathop {\lim }\limits_{x \to \infty } \frac{(1 - a^2)x - 1 + \frac{1}{x}}{\sqrt{1 - \frac{1}{x} + \frac{1}{x^2}} + a} = b$

For this limit to be finite, the term $(1-a^2)x$ in the numerator must vanish. This implies: $1 - a^2 = 0 \implies a^2 = 1 \implies a = \pm 1$ As reasoned in Step 1, for a finite limit, we must have $a=1$.

Now substitute $a=1$ into the limit expression: $\mathop {\lim }\limits_{x \to \infty } \frac{(1 - 1^2)x - 1 + \frac{1}{x}}{\sqrt{1 - \frac{1}{x} + \frac{1}{x^2}} + 1} = b$ $\mathop {\lim }\limits_{x \to \infty } \frac{0 \cdot x - 1 + \frac{1}{x}}{\sqrt{1 - \frac{1}{x} + \frac{1}{x^2}} + 1} = b$ $\mathop {\lim }\limits_{x \to \infty } \frac{-1 + \frac{1}{x}}{\sqrt{1 - \frac{1}{x} + \frac{1}{x^2}} + 1} = b$

Step 6: Calculate the Value of 'b' (Revised) Now, as $x \to \infty$, $\frac{1}{x} \to 0$ and $\frac{1}{x^2} \to 0$. $b = \frac{-1 + 0}{\sqrt{1 - 0 + 0} + 1}$ $b = \frac{-1}{\sqrt{1} + 1}$ $b = \frac{-1}{1 + 1}$ $b = \frac{-1}{2}$

So, we have found $a=1$ and $b = -\frac{1}{2}$. The ordered pair is $(1, -\frac{1}{2})$.

Common Mistakes & Tips

• Incorrectly identifying the indeterminate form: Always check if the expression leads to $\infty - \infty$, $0/0$, or $\infty/\infty$ before applying techniques like rationalization.
• Errors in algebraic manipulation: Be very careful with signs and exponents when expanding squares and simplifying fractions.
• Forgetting to divide the entire numerator/denominator by the highest power of x: This is a common mistake when dealing with limits at infinity involving square roots. Ensure you divide all terms correctly.
• Assuming $a=-1$ is valid: While $a^2=1$ allows for $a=-1$, this would lead to a different limit behavior. For the limit to be finite, we must have $a=1$.

Summary

The problem requires evaluating a limit of the form $\infty - \infty$. We first rationalize the expression to transform it into a form that can be evaluated. By analyzing the dominant terms in the numerator and denominator after rationalization, we determine that for the limit to be finite, the coefficient of the $x^2$ term in the numerator must be zero, which leads to $a=1$. Substituting $a=1$ back into the limit expression and evaluating as $x \to \infty$ allows us to find the value of $b$.

The ordered pair $(a, b)$ is $(1, -1/2)$.

The final answer is \boxed{\left( {1, - {1 \over 2}} \right)}. This corresponds to option (B).