If <span class="math-inline">\( \lim_{x \to \infty} \left( \frac{x^2 + x + 1}{x + 1} - ax - b \right) = 4 \)</span>, then -
Step-by-Step Solution
Key Concept: For the limit to exist and equal a finite value as x→∞, the coefficient of the highest degree term in the expression must vanish. We must expand the rational function, subtract the linear terms, and match coefficients to make the limit finite.
<p><strong>Step 1:</strong> Perform polynomial long division on $\frac{x^2 + x + 1}{x + 1}$.</p><p>Divide: $x^2 + x + 1 = (x+1) \cdot q(x) + r$</p><p>$x^2 + x + 1 = (x+1)(x) + 1$</p><p>So $\frac{x^2 + x + 1}{x + 1} = x + \frac{1}{x+1}$</p><p><strong>Step 2:</strong> Substitute this into the limit expression.</p><p>$$\lim_{x \to \infty} \left( x + \frac{1}{x+1} - ax - b \right) = 4$$</p><p><strong>Step 3:</strong> Rearrange to group terms by powers of x.</p><p>$$\lim_{x \to \infty} \left( (1-a)x + \frac{1}{x+1} - b \right) = 4$$</p><p><strong>Step 4:</strong> For this limit to exist and be finite as $x \to \infty$, the coefficient of x must be zero.</p><p>Therefore: $1 - a = 0 \Rightarrow a = 1$</p><p><strong>Step 5:</strong> With $a = 1$, evaluate the remaining limit.</p><p>$$\lim_{x \to \infty} \left( \frac{1}{x+1} - b \right) = 4$$</p><p>As $x \to \infty$, $\frac{1}{x+1} \to 0$</p><p>Therefore: $0 - b = 4 \Rightarrow b = -4$</p><p><strong>Verification:</strong> With $a=1$ and $b=-4$: $\lim_{x \to \infty} \left( x + \frac{1}{x+1} - x + 4 \right) = \lim_{x \to \infty} \left( 4 + \frac{1}{x+1} \right) = 4$ ✓</p><p><strong>∴ Answer:</strong> B</p>
Correct Answer: B