Limits, Continuity & Differentiability
Limits of the form 1^infinity
Grade 12
<p>Evaluate: \(\lim_{x \to 0} \left(\frac{\sin x}{x}\right)^{\frac{\sin x}{x - \sin x}}\)</p>
Step-by-Step Solution
Key Concept: Rewrite the exponent using the algebraic identity a/(a-b) = 1 + b/(a-b), then recognize that as x→0, the exponent approaches -1 while the base approaches 1, creating a 1^∞ indeterminate form requiring logarithmic analysis.
<p><strong>Step 1:</strong> Let L = lim_{x→0} (sin x/x)^(sin x/(x - sin x)). Take natural log: ln L = lim_{x→0} (sin x/(x - sin x)) · ln(sin x/x)</p><p><strong>Step 2:</strong> Rewrite the exponent: sin x/(x - sin x) = sin x/[x(1 - sin x/x)] = (sin x/x)/(1 - sin x/x) = a/(1-a) where a = sin x/x. As x→0, a→1, so the exponent → 1/(1-1), which is problematic. Instead, manipulate: sin x/(x - sin x) = 1/[(x - sin x)/sin x] = 1/[x/sin x - 1]. Since lim_{x→0} x/sin x = 1, we have x/sin x - 1 → 0.</p><p><strong>Step 3:</strong> Better approach: Write sin x/(x - sin x) = sin x/x · 1/(1 - sin x/x) = -1 · (sin x/x)/(sin x/x - 1). As x→0, sin x/x → 1, so let sin x/x = 1 + u where u→0. Then exponent = -(1+u)/u = -1/u - 1 → -1 as the dominant term.</p><p><strong>Step 4:</strong> More rigorously: ln L = lim_{x→0} [sin x/(x - sin x)] · ln(sin x/x) = lim_{x→0} [sin x/(x - sin x)] · [(sin x - x)/x] = lim_{x→0} sin x(sin x - x)/[x(x - sin x)] = lim_{x→0} -sin x/x = -1</p><p><strong>Step 5:</strong> Therefore ln L = -1, which gives L = e^{-1}</p><p>∴ Answer: <strong>e^{-1}</strong></p>
Correct Answer: e^{-1}