Limits, Continuity & Differentiability
Existence of finite limits
Grade 12
<p>If \(\lim_{x \to 0} \dfrac{ae^x + b\sin 2x + c\sqrt{1-x}}{x^2}\) exists finitely, then:</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, linear term, and quadratic term coefficients in the Taylor expansion of the numerator to all equal zero.
<p><strong>Step 1: Expand each term using Taylor series about x=0</strong></p><p>• $e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + ...$</p><p>• $\sin 2x = 2x - \frac{(2x)^3}{6} + ... = 2x - \frac{4x^3}{3} + ...$</p><p>• $\sqrt{1-x} = 1 - \frac{x}{2} - \frac{x^2}{8} - \frac{x^3}{16} - ...$</p><p><strong>Step 2: Write numerator as</strong></p><p>$ae^x + b\sin 2x + c\sqrt{1-x}$</p><p>$= a(1 + x + \frac{x^2}{2} + ...) + b(2x - \frac{4x^3}{3} + ...) + c(1 - \frac{x}{2} - \frac{x^2}{8} - ...)$</p><p><strong>Step 3: Collect coefficients by powers of x</strong></p><p>• Constant term: $a + c = 0$ → $c = -a$</p><p>• Coefficient of $x$: $a + 2b - \frac{c}{2} = 0$</p><p>• Coefficient of $x^2$: $\frac{a}{2} - \frac{c}{8} = 0$</p><p><strong>Step 4: Solve the system</strong></p><p>From Step 3: $c = -a$ and $\frac{a}{2} - \frac{(-a)}{8} = \frac{a}{2} + \frac{a}{8} = \frac{5a}{8} = 0$ → $a = 0$</p><p>Therefore: $a = 0$, $c = 0$, and from $0 + 2b - 0 = 0$ → $b = 0$</p><p>∴ Answer: A (The conditions are a=0, b=0, c=0, or equivalent relationship)</p>
Correct Answer: A