Limits, Continuity & Differentiability
Continuity of functions
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 left and right limits must equal f(0). Use Taylor expansions of sin x and cos x to find the coefficients that make the numerator vanish to the required order at x=0.
<p><strong>Step 1:</strong> For continuity at x=0, we need: lim(x→0⁻) f(x) = lim(x→0⁺) f(x) = f(0).</p><p><strong>Step 2:</strong> For x<0, f(x) = cx + d. So lim(x→0⁻) f(x) = d, which means f(0) = d.</p><p><strong>Step 3:</strong> For x>0, expand using Taylor series: sin x = x - x³/6 + ... and cos x = 1 - x²/2 + ...</p><p><strong>Step 4:</strong> Numerator = a(1 - x(x - x³/6 + ...)) + b(1 - x²/2 + ...) + 5<br/>= a(1 - x² + ...) + b(1 - x²/2 + ...) + 5<br/>= (a + b + 5) - ax² - bx²/2 + ...</p><p><strong>Step 5:</strong> For the limit to exist and equal d: constant term must equal d, so a + b + 5 = d, and the x² coefficient must equal 0, so -a - b/2 = 0, giving b = -2a.</p><p><strong>Step 6:</strong> From a + b + 5 = d with b = -2a: a - 2a + 5 = d, so d = 5 - a.</p><p><strong>Step 7:</strong> This gives us the constraints: <strong>a + b + 5 = d</strong> and <strong>2a + b = 0</strong>.</p><p>∴ Answer: AB</p>
Correct Answer: AB