Limits, Continuity & Differentiability
Limits
Grade 12
Question:
<p>If \(\lim_{x \to \infty} \left(1 + \dfrac{a}{x} - \dfrac{4}{x^2}\right)^{2x} = e^3\), then \(a\) is equal to</p>
<p>2</p>
<p>\(\dfrac{3}{2}\)</p>
<p>\(\dfrac{1}{2}\)</p>
<p>\(\dfrac{2}{3}\)</p>
Step-by-Step Solution
Key Concept: Rewrite the expression in the standard form (1 + u)^(1/u) where u → 0, then match exponents using logarithms to find a. The limit e^3 tells us that the exponent of e must equal 3.
<p><strong>Step 1:</strong> Recognize the standard form. For limits of type (1 + u)^(1/u) as u → 0, we have lim = e.</p><p><strong>Step 2:</strong> Rewrite the given expression:</p><p>$$\lim_{x \to \infty} \left(1 + \frac{a}{x} - \frac{4}{x^2}\right)^{2x}$$</p><p><strong>Step 3:</strong> Take the natural logarithm of the limit:</p><p>$$\ln(e^3) = \lim_{x \to \infty} 2x \ln\left(1 + \frac{a}{x} - \frac{4}{x^2}\right)$$</p><p><strong>Step 4:</strong> Apply the expansion $\ln(1 + t) \approx t - \frac{t^2}{2} + ...$ where $t = \frac{a}{x} - \frac{4}{x^2}$:</p><p>$$\ln\left(1 + \frac{a}{x} - \frac{4}{x^2}\right) \approx \frac{a}{x} - \frac{4}{x^2} - \frac{1}{2}\left(\frac{a}{x}\right)^2 + ...$$</p><p><strong>Step 5:</strong> Multiply by 2x:</p><p>$$2x \ln\left(1 + \frac{a}{x} - \frac{4}{x^2}\right) \approx 2a - \frac{8}{x} - \frac{a^2}{x} + ... \to 2a$$</p><p><strong>Step 6:</strong> Equate with the exponent:</p><p>$$2a = 3$$</p><p>$$a = \frac{3}{2}$$</p><p>∴ Answer: A</p>
Correct Answer: A