Key Concepts and Formulas

• Limit of a Function: Understanding how to evaluate limits, especially indeterminate forms like $\frac{0}{0}$ or $\infty - \infty$.
• Algebraic Manipulation of Limits: Techniques like multiplying by the conjugate to resolve indeterminate forms involving square roots.
• Dominant Terms in Limits: Identifying the highest power of the variable that dictates the limit's behavior as the variable approaches infinity.
• Binomial Expansion (for large n): The approximation $\sqrt{1 - x} \approx 1 - \frac{1}{2}x$ for small $x$, or more generally, extracting $n$ from the square root.

Step-by-Step Solution

Step 1: Analyze the Limit and Identify the Indeterminate Form We are given the limit: $\mathop {\lim }\limits_{n \to \infty } \left( {\sqrt {{n^2} - n - 1} + n\alpha + \beta } \right) = 0$ As $n \to \infty$, $\sqrt{n^2 - n - 1}$ behaves like $\sqrt{n^2} = n$. So, the expression inside the limit becomes approximately $n + n\alpha + \beta$. For this limit to be 0, the dominant terms involving $n$ must cancel out. If $\alpha > -1$, the term $n\alpha$ would lead to an overall limit of $\infty$. If $\alpha < -1$, the limit would be $-\infty$. Therefore, for the limit to be finite (and in this case, 0), we must have $\alpha = -1$. This ensures that the terms involving $n$ cancel. Let's rewrite the expression with $\alpha = -1$: $\sqrt{n^2 - n - 1} - n + \beta$ As $n \to \infty$, this is an indeterminate form of type $\infty - \infty$.

Step 2: Resolve the Indeterminate Form using Conjugate Multiplication To resolve the $\infty - \infty$ form, we multiply by the conjugate of the expression. $\sqrt{n^2 - n - 1} - n + \beta = \left( {\sqrt {{n^2} - n - 1} - n} \right) + \beta$ Let's focus on the part involving the square root first: $\left( {\sqrt {{n^2} - n - 1} - n} \right) = \frac{\left( {\sqrt {{n^2} - n - 1} - n} \right)\left( {\sqrt {{n^2} - n - 1} + n} \right)}{\sqrt {{n^2} - n - 1} + n}$ $= \frac{(n^2 - n - 1) - n^2}{\sqrt {{n^2} - n - 1} + n} = \frac{-n - 1}{\sqrt {{n^2} - n - 1} + n}$ Now, we can divide the numerator and denominator by $n$ to evaluate the limit of this part: $\mathop {\lim }\limits_{n \to \infty } \frac{-n - 1}{\sqrt {{n^2} - n - 1} + n} = \mathop {\lim }\limits_{n \to \infty } \frac{-1 - \frac{1}{n}}{\sqrt {1 - \frac{1}{n} - \frac{1}{n^2}} + 1} = \frac{-1 - 0}{\sqrt {1 - 0 - 0} + 1} = \frac{-1}{1 + 1} = -\frac{1}{2}$ So, the original limit becomes: $\mathop {\lim }\limits_{n \to \infty } \left( {\left( {\sqrt {{n^2} - n - 1} - n} \right) + \beta } \right) = -\frac{1}{2} + \beta$ We are given that this limit is equal to 0. $-\frac{1}{2} + \beta = 0$ $\beta = \frac{1}{2}$

Step 3: Verify the Condition on $\alpha$ Our initial analysis in Step 1 concluded that $\alpha = -1$ is necessary for the limit to be finite. The original solution's reasoning about the powers of $n$ in the numerator and denominator after multiplying by the conjugate is a valid way to arrive at this conclusion as well, by ensuring the dominant $n^2$ term in the numerator cancels out.

Let's re-trace the algebraic steps from the original solution to confirm: The expression after multiplying by the conjugate is: $\frac{{\left( {{n^2} - n - 1} \right) - {{\left( {n\alpha + \beta } \right)}^2}}}{{\sqrt {{n^2} - n - 1} - \left( {n\alpha + \beta } \right)}} = \frac{{n^2 - n - 1 - (n^2\alpha^2 + 2n\alpha\beta + \beta^2)}}{{\sqrt {{n^2} - n - 1} - \left( {n\alpha + \beta } \right)}}$ $= \frac{{n^2(1 - \alpha^2) - n(1 + 2\alpha\beta) - (1 + \beta^2)}}{{\sqrt {{n^2} - n - 1} - \left( {n\alpha + \beta } \right)}}$ For the limit to be 0, as $n \to \infty$, the numerator must not grow faster than the denominator. The highest power of $n$ in the denominator is $n^1$ (from $\sqrt{n^2}$). For the limit to be finite, the $n^2$ term in the numerator must be zero. $1 - \alpha^2 = 0 \implies \alpha^2 = 1 \implies \alpha = \pm 1$ If $\alpha = 1$, the expression becomes $\sqrt{n^2 - n - 1} + n + \beta$, which tends to $\infty$. If $\alpha = -1$, the expression becomes $\sqrt{n^2 - n - 1} - n + \beta$, which is the $\infty - \infty$ form we analyzed. Thus, $\alpha = -1$.

Now, substitute $\alpha = -1$ into the numerator: $n^2(1 - (-1)^2) - n(1 + 2(-1)\beta) - (1 + \beta^2) = n^2(0) - n(1 - 2\beta) - (1 + \beta^2)$ $= -n(1 - 2\beta) - (1 + \beta^2)$ The limit expression becomes: $\mathop {\lim }\limits_{n \to \infty } \frac{-n(1 - 2\beta) - (1 + \beta^2)}{\sqrt {{n^2} - n - 1} - (-n + \beta)} = \mathop {\lim }\limits_{n \to \infty } \frac{-n(1 - 2\beta) - (1 + \beta^2)}{\sqrt {{n^2} - n - 1} + n - \beta}$ Divide numerator and denominator by $n$: $\mathop {\lim }\limits_{n \to \infty } \frac{-(1 - 2\beta) - \frac{1 + \beta^2}{n}}{\sqrt {1 - \frac{1}{n} - \frac{1}{n^2}} + 1 - \frac{\beta}{n}}$ As $n \to \infty$, this simplifies to: $\frac{-(1 - 2\beta)}{\sqrt{1} + 1} = \frac{-(1 - 2\beta)}{2}$ We are given that the limit is 0: $\frac{-(1 - 2\beta)}{2} = 0$ $-(1 - 2\beta) = 0$ $1 - 2\beta = 0$ $2\beta = 1$ $\beta = \frac{1}{2}$ This confirms our values: $\alpha = -1$ and $\beta = \frac{1}{2}$.

Step 4: Calculate the Final Expression We need to find the value of $8(\alpha + \beta)$. $8(\alpha + \beta) = 8\left( {-1 + \frac{1}{2}} \right)$ $= 8\left( {-\frac{1}{2}} \right)$ $= -4$

Common Mistakes & Tips

• Incorrectly handling the indeterminate form: Recognizing that $\infty - \infty$ requires algebraic manipulation (like multiplying by the conjugate) is crucial. Simply substituting values will not work.
• Errors in algebraic simplification: Be very careful with signs and powers when expanding and rearranging terms, especially when squaring $(n\alpha + \beta)$.
• Confusing $n \to \infty$ with $n \to \alpha$: The problem states $n \to \infty$, not $n \to \alpha$. The original solution incorrectly used $n \to \alpha$ in several steps. This mistake needs to be corrected for a proper derivation. The limit is taken as $n \to \infty$.

Summary

The problem requires evaluating a limit that results in an indeterminate form of $\infty - \infty$. This is resolved by first identifying the necessary condition for the limit to be finite, which implies $\alpha = -1$. Then, by multiplying the expression by its conjugate, we transform the limit into a form where the highest power of $n$ in the numerator cancels out, allowing us to further simplify and find the value of $\beta$. Once $\alpha$ and $\beta$ are determined, we compute the required expression $8(\alpha + \beta)$.

The final answer is \boxed{-4}.