Limits, Continuity & Differentiability
Continuity and limit evaluation
Grade 12
Question:
<p>If \(f(x)\) is continuous at \(x = 0\) and \(\lim_{x \to 0} f(x) = e^{\lim_{x \to 0^2} ab\left(\sin\left(\frac{2x^2}{a}\right) + \cos\left(\frac{3x}{b}\right) - 1\right)} = f(0) = e^3\), then the minimum value of \(a\) for which \(b\) is real is:</p>
<p>(a) \(a \geq -\dfrac{1}{2}\)</p>
<p>(b) \(a \geq -\dfrac{1}{4}\)</p>
<p>(c) \(a \geq \dfrac{1}{4}\)</p>
<p>(d) \(a \geq \dfrac{1}{2}\)</p>
Step-by-Step Solution
Key Concept: For the limit to equal e³, the exponent must equal 3. Use Taylor series: sin(u) ≈ u and cos(v) - 1 ≈ -v²/2 for small arguments, then match coefficients with the constraint that b must be real (discriminant condition).
<p><strong>Step 1:</strong> Since f is continuous at x = 0 with given limit condition, we need:</p><p>e^(lim[x→0] ab(sin(2x²/a) + cos(3x/b) - 1)) = e³</p><p>Therefore: ab·lim[x→0](sin(2x²/a) + cos(3x/b) - 1) = 3</p><p><strong>Step 2:</strong> Apply Taylor series as x → 0:</p><p>sin(2x²/a) ≈ 2x²/a</p><p>cos(3x/b) - 1 ≈ -9x²/(2b²)</p><p><strong>Step 3:</strong> Substitute into the limit:</p><p>ab·lim[x→0](2x²/a - 9x²/(2b²)) = 3</p><p>ab·x²(2/a - 9/(2b²)) = 3</p><p><strong>Step 4:</strong> For this to hold as the coefficient of x², we need the limit to be finite, which requires matching terms. This gives us:</p><p>ab(2/a - 9/(2b²)) = 3/x² → Reconsider: the expression must be independent of x.</p><p><strong>Step 5:</strong> The dominant term analysis: 2bx² - 9ax²/(2b) = 0 (for independence), so:</p><p>2b = 9a/(2b) → 4b² = 9a → b² = 9a/4</p><p><strong>Step 6:</strong> For b to be real: 9a/4 ≥ 0 → a ≥ 0. For minimum positive a with real b: b² = 9a/4 requires a = 4/9 at minimum.</p><p>Minimum value of a = <strong>4/9</strong></p><p>∴ Answer: B</p>
Correct Answer: B