<p>\(\lim_{x \to \infty} \left(\dfrac{x^2 + 5x + 3}{x^2 + x + 2}\right)^x\) is equal to</p>
Step-by-Step Solution
Key Concept: Rewrite the fraction as (1 + ratio of differences) and use the standard limit form (1 + 1/n)^n → e. The key is extracting the dominant behavior by dividing numerator and denominator differences by the denominator itself.
<p><strong>Step 1:</strong> Rewrite the fraction inside the parentheses.</p><p>$$\frac{x^2 + 5x + 3}{x^2 + x + 2} = \frac{(x^2 + x + 2) + (4x + 1)}{x^2 + x + 2} = 1 + \frac{4x + 1}{x^2 + x + 2}$$</p><p><strong>Step 2:</strong> Apply the exponential limit form: $\lim_{x \to \infty} \left(1 + \frac{4x + 1}{x^2 + x + 2}\right)^x$</p><p><strong>Step 3:</strong> For the exponent, multiply and divide strategically:</p><p>$$= \lim_{x \to \infty} \left[\left(1 + \frac{4x + 1}{x^2 + x + 2}\right)^{\frac{x^2+x+2}{4x+1}}\right]^{\frac{(4x+1)x}{x^2+x+2}}$$</p><p><strong>Step 4:</strong> The inner limit equals $e$ as $x \to \infty$ (standard form).</p><p><strong>Step 5:</strong> Evaluate the exponent:</p><p>$$\lim_{x \to \infty} \frac{(4x+1)x}{x^2+x+2} = \lim_{x \to \infty} \frac{4x^2 + x}{x^2 + x + 2} = \lim_{x \to \infty} \frac{4 + 1/x}{1 + 1/x + 2/x^2} = 4$$</p><p>∴ Answer: $e^4$</p>
Correct Answer: A