Limits, Continuity & Differentiability
Limits of nth Roots
Grade 12
Question:
<p>If \(\lim_{n \to \infty} \frac{\sqrt[n]{(2n)^n}}{n}\) is equals to \((a-1)e^b\), then</p>
<p>(A) \(a = 2\)</p>
<p>(B) \(a = 1\)</p>
<p>(C) \(b = -1\)</p>
<p>(D) \(b = 1\)</p>
Step-by-Step Solution
Key Concept: Simplify nth roots and identify the limiting value by recognizing that $\sqrt[n]{x^n} = x$.
<p><strong>Step 1:</strong> Simplify $\sqrt[n]{(2n)^n} = 2n$.</p><p><strong>Step 2:</strong> $\frac{\sqrt[n]{(2n)^n}}{n} = \frac{2n}{n} = 2$.</p><p><strong>Step 3:</strong> Therefore $\lim_{n \to \infty} \frac{\sqrt[n]{(2n)^n}}{n} = 2$.</p><p><strong>Step 4:</strong> Setting $2 = (a-1)e^b$, we get $a = 2, b = 0$ as one solution, or $a = 2$ as the answer.</p>
Correct Answer: A