<p>\(\lim_{x \to \infty} \sum_{r=1}^{n} \dfrac{1}{n} e^{r/n}\) is</p>
Step-by-Step Solution
Key Concept: Recognize this sum as a Riemann sum approximation where the limit corresponds to the definite integral ∫₀¹ eˣ dx. Here, Δx = 1/n and the function is evaluated at points r/n.
<p><strong>Step 1:</strong> Identify the Riemann sum structure. We have ∑(r=1 to n) (1/n)·e^(r/n), where:</p><ul><li>Width of each subinterval: Δx = 1/n</li><li>Number of subintervals: n</li><li>Function: f(x) = eˣ</li><li>Evaluation points: xᵣ = r/n for r = 1, 2, ..., n</li></ul><p><strong>Step 2:</strong> Recognize this as a right Riemann sum for f(x) = eˣ on [0,1] as n→∞:</p><p>lim(n→∞) ∑(r=1 to n) (1/n)·e^(r/n) = ∫₀¹ eˣ dx</p><p><strong>Step 3:</strong> Evaluate the definite integral:</p><p>∫₀¹ eˣ dx = [eˣ]₀¹ = e¹ - e⁰ = e - 1</p><p><strong>∴ Answer: e - 1 (Option B)</strong></p>
Correct Answer: B