<p>For \(x \in \mathbb{R}\), \(\lim_{x \to \infty} \left(\dfrac{x-3}{x+2}\right)^x\) is equal to</p>
Step-by-Step Solution
Key Concept: Rewrite the fraction as 1 + (a small term) to use the standard limit form lim(1 + 1/n)^n = e. Here, (x-3)/(x+2) = 1 - 5/(x+2), so the exponent times the fractional part gives -5.
<p><strong>Step 1:</strong> Recognize this is the indeterminate form 1^∞. Rewrite the base:</p><p>$$\frac{x-3}{x+2} = \frac{x+2-5}{x+2} = 1 - \frac{5}{x+2}$$</p><p><strong>Step 2:</strong> Use the standard limit form. Set the exponent as follows:</p><p>$$\lim_{x \to \infty} \left(1 - \frac{5}{x+2}\right)^x = \lim_{x \to \infty} \left[\left(1 - \frac{5}{x+2}\right)^{(x+2)/(-5)}\right]^{-5x/(x+2)}$$</p><p><strong>Step 3:</strong> As x → ∞, the inner bracket approaches e (by standard limit), and the outer exponent approaches:</p><p>$$\lim_{x \to \infty} \frac{-5x}{x+2} = -5$$</p><p><strong>Step 4:</strong> Therefore:</p><p>$$\lim_{x \to \infty} \left(\frac{x-3}{x+2}\right)^x = e^{-5}$$</p><p>∴ Answer: C (e^{-5})</p>
Correct Answer: C