Limits, Continuity & Differentiability
Continuity and limit evaluation
Grade 12
Question:
<p>Let a function is defined as \[f(x) = \begin{cases} \dfrac{a(1 - x\sin x) + b\cos x + 5}{x^2} & \text{if } x < 0 \\ 3 & \text{if } x = 0 \\ \left(1 + \left(\dfrac{cx + dx^3}{x^2}\right)\right)^{1/x} & \text{if } x > 0 \end{cases}\] where \(a, b, c\) and \(d\) be constants, if \(f(x)\) is continuous at \(x = 0\), then:</p>
<p>(a) \(\lim_{x \to 0} \dfrac{e^{bx} - 1}{\sin x} = -4\)</p>
<p>(b) number of points of discontinuity of \(g(x) = [c - 3a\sin x]\) in \([0, \pi]\) is 5.</p>
Step-by-Step Solution
Key Concept: For f(x) to be continuous at x=0, the limit as x→0 of the piecewise expression must equal f(0). Use Taylor series expansions of sin x and cos x to evaluate the limit of the rational function, then match coefficients to find constraints on a and b.
<p><strong>Step 1:</strong> For continuity at x=0, we need lim(x→0) f(x) = f(0) = c</p><p><strong>Step 2:</strong> Expand the numerator using Taylor series: sin x = x - x³/6 + ..., cos x = 1 - x²/2 + ...</p><p><strong>Step 3:</strong> Substitute: a(1 - x·(x - x³/6 + ...)) + b(1 - x²/2 + ...) + 5</p><p>= a(1 - x² + x⁴/6 + ...) + b(1 - x²/2 + ...) + 5</p><p>= (a + b + 5) + (-a - b/2)x² + (higher order terms)</p><p><strong>Step 4:</strong> Divide by x²: The expression becomes [a + b + 5]/x² + (-a - b/2) + ..., which only has a finite limit if a + b + 5 = 0</p><p><strong>Step 5:</strong> When a + b + 5 = 0, the limit equals (-a - b/2) = (a + b)/2 + b/2 = -5/2 + b/2</p><p><strong>Step 6:</strong> Setting a + b = -5 and c = lim(x→0) f(x), we get constraints: <strong>a + b + 5 = 0 and c = d</strong></p><p>∴ Answer: AB</p>
Correct Answer: AB