Limits, Continuity & Differentiability
Limits
Grade 12
Question:
<p>If \(\lim_{x \to \infty} \left(1 + \dfrac{a}{x} + \dfrac{b}{x^2}\right)^{2x} = e^2\), then the values of \(a\) and \(b\), are</p>
<p>\(a \in \mathbb{R},\ b \in \mathbb{R}\)</p>
<p>\(a = 1,\ b \in \mathbb{R}\)</p>
<p>\(a \in \mathbb{R},\ b = 2\)</p>
<p>\(a = 1\) and \(b = 2\)</p>
Step-by-Step Solution
Key Concept: Recognize this as a standard indeterminate form (1^∞) and use the expansion (1 + u)^n ≈ e^(nu) when u → 0, combined with the requirement that the exponent coefficient must match e^2.
<p><strong>Step 1:</strong> Take logarithm of the expression:</p><p>Let L = lim_{x → ∞} (1 + a/x + b/x²)^(2x)</p><p>ln(L) = lim_{x → ∞} 2x · ln(1 + a/x + b/x²)</p><p><strong>Step 2:</strong> Use Taylor expansion ln(1 + u) = u - u²/2 + ... where u = a/x + b/x²:</p><p>ln(1 + a/x + b/x²) = (a/x + b/x²) - (a/x + b/x²)²/2 + ...</p><p>= a/x + b/x² - a²/(2x²) + O(1/x³)</p><p><strong>Step 3:</strong> Multiply by 2x:</p><p>2x · ln(1 + a/x + b/x²) = 2a + 2b/x - a²/x + O(1/x²)</p><p><strong>Step 4:</strong> Take limit as x → ∞:</p><p>ln(L) = 2a</p><p>Therefore: L = e^(2a)</p><p><strong>Step 5:</strong> Compare with given limit e²:</p><p>e^(2a) = e²</p><p>2a = 2 ⟹ <strong>a = 1</strong></p><p><strong>Step 6:</strong> For the limit to exist and be finite, the coefficient of 1/x must vanish:</p><p>2b - a² = 0</p><p>2b - 1 = 0 ⟹ <strong>b = 1/2</strong></p><p>∴ Answer: a = 1, b = 1/2</p>
Correct Answer: B