Limits, Continuity & Differentiability
Existence of finite limits
Grade 12
Question:
<p>If \(\displaystyle\lim_{x\to 0}\frac{ae^x + b\sin 2x + c\sqrt{1-x}}{x^2}\) exists finitely, then:</p>
<p>\(a + c = 0\)</p>
<p>\(2a + 4b - c = 0\)</p>
<p>\(2a + 3b = 0\)</p>
<p>\(3a + 4b = 0\)</p>
Step-by-Step Solution
Key Concept: For a limit of the form f(x)/x² to exist finitely as x→0, the numerator must have a zero of at least order 2 at x=0. This requires the constant term and coefficient of x to both vanish, giving us two equations in three unknowns.
<p><strong>Step 1: Expand each term using Taylor series as x→0</strong></p><p>• e^x = 1 + x + x²/2 + x³/6 + ...</p><p>• sin 2x = 2x - (2x)³/3! + ... = 2x - 4x³/3 + ...</p><p>• √(1-x) = 1 - x/2 - x²/8 - x³/16 + ...</p><p><strong>Step 2: Substitute into numerator</strong></p><p>Numerator = a(1 + x + x²/2 + ...) + b(2x - 4x³/3 + ...) + c(1 - x/2 - x²/8 - ...)</p><p>= (a + c) + (a + 2b - c/2)x + (a/2 - c/8)x² + ...</p><p><strong>Step 3: Apply condition for finite limit</strong></p><p>For lim[numerator/x²] to exist finitely, we need:</p><p>• Constant term: a + c = 0 → <strong>a = -c</strong></p><p>• Coefficient of x: a + 2b - c/2 = 0</p><p>Substituting a = -c: -c + 2b - c/2 = 0 → 2b = 3c/2 → <strong>b = 3c/4</strong></p><p><strong>Step 4: Express relationships</strong></p><p>We have: a = -c and b = 3c/4, or equivalently: 4a + 4b + 4c = 0 (one relation) and specific ratios between a, b, c.</p><p>∴ Answer: The relationships a = -c and 4b = 3c (or 4a + 4b + 4c = 0) must hold</p>
Correct Answer: ABC