If lim(n→∞) (√(n² - n - 1) + nα + β) = 0 then 8(α + β) is equal to :

Q: Why can't we just say $\sqrt{n^2 - n - 1} \approx n$ and solve $n + n\alpha + \beta = 0$?

That's a common trap! While $\sqrt{n^2 - n - 1}$ does behave like $n$ for very large $n$, the *next* term in the expansion is crucial. The $-1/2$ term is the difference between a finite limit and infinity. You must expand to at least the $1/n$ term to capture the constant part.

Q: Why do we set the coefficient of $n$ to zero?

If the coefficient of $n$ were anything other than zero, the expression would grow to $\pm \infty$ as $n \to \infty$. Since the limit is given as $0$, the expression must be bounded. The only way to ensure this is to force the diverging part to vanish.

Q: Can I use L'Hopital's Rule here?

You could, but it's like using a sledgehammer to crack a nut. You'd have to rewrite the expression as a fraction, which gets messy with the square root. Binomial expansion is the elegant, professional way to handle limits at infinity involving radicals.

Question

If $\lim_{n 	o \infty} (\sqrt{n^2 - n - 1} + n\alpha + \beta) = 0$ then $8(\alpha + \beta)$ is equal to :

Options:

(A) 4

(B) -8

(C) -4

(D) 8

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the edge of an infinite horizon. You are looking at the expression:

$$\lim_{n 	o \infty} (\sqrt{n^2 - n - 1} + n\alpha + \beta) = 0$$

This is a classic balancing act. A square root that grows without bound is added to a linear term, and they are forced to settle down to zero.

### The Indeterminate Trap

First, let's look at the behavior of $\sqrt{n^2 - n - 1}$ as $n 	o \infty$. It clearly grows towards infinity.

If $\alpha$ were positive, the term $n\alpha$ would also grow to infinity, and the sum would explode. Since the limit is zero, $\alpha$ must be negative to create an $\infty - \infty$ indeterminate form. We are essentially forcing two opposing infinities to cancel each other out perfectly.

### The Binomial Weapon

To resolve this, we factor out $n^2$ from the square root:

$$\sqrt{n^2(1 - \frac{1}{n} - \frac{1}{n^2})} = n(1 - \frac{1}{n} - \frac{1}{n^2})^{\frac{1}{2}}$$

Now, we apply the binomial expansion $(1+x)^k \approx 1 + kx$. Here, $x = -\frac{1}{n} - \frac{1}{n^2}$ and $k = \frac{1}{2}$.

This yields:

$$n(1 - \frac{1}{2n} - \frac{1}{2n^2} + \dots) = n - \frac{1}{2} - \frac{1}{2n} + \dots$$

This expansion is the key, as it isolates the constant term $-\frac{1}{2}$.

### The Balancing Act

Now, our expression becomes:

$$n - \frac{1}{2} - \frac{1}{2n} + n\alpha + \beta$$

Grouping the terms by their power of $n$, we get:

$$n(1 + \alpha) + (\beta - \frac{1}{2}) - \frac{1}{2n} + \dots = 0$$

For the limit to be zero, the coefficient of $n$ must be zero:

$$1 + \alpha = 0 \Rightarrow \alpha = -1$$

With the $n$ term eliminated, the remaining constant part must also be zero:

$$\beta - \frac{1}{2} = 0 \Rightarrow \beta = \frac{1}{2}$$

### The Final Victory

We have determined the constants: $\alpha = -1$ and $\beta = \frac{1}{2}$. The problem asks for the value of $8(\alpha + \beta)$.

Substituting our values:

$$8(-1 + \frac{1}{2}) = 8(-\frac{1}{2}) = -4$$

The final result is **-4**. Remember, when you encounter limits at infinity, do not panic; simply expand, balance, and solve.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $lim_{n \to \infty} (n^{2} - n - 1 + n α + β) = 0$ then $8 (α + β)$ is equal to :

Select Answer:

Visualized Solution

Analyze the Limit Form

Identify the Indeterminate Form

Factoring $n^{2}$ from the Radical

Applying Binomial Expansion

Simplifying the Expansion

Grouping Terms by Powers of $n$

Constraint for Finite Limit

Solving for $α$

Constraint for Zero Limit

Solving for $β$

Final Calculation Setup

Atomic Compute: Final Result

Summary and Key Takeaway

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Setup

The Indeterminate Trap

The Binomial Weapon

The Balancing Act

The Final Victory

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Animated Solution for Mathematics - Limits, Continuity and Differentiability: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If limn→∞​(n2−n−1​+nα+β)=0 then 8(α+β) is equal to :