<p>Evaluate <span>\(\lim_{x \to 0^+} (\sin x)^{\tan x}\)</span></p>
Step-by-Step Solution
Key Concept: For indeterminate forms like 0^0, use logarithmic transformation: if y = (sin x)^(tan x), then ln y = tan x · ln(sin x), and evaluate the limit of ln y. This converts the problem into a 0·∞ indeterminate form that can be resolved using L'Hôpital's rule.
<p><strong>Step 1:</strong> Recognize the indeterminate form. As x → 0⁺, sin x → 0⁺ and tan x → 0⁺, giving the form 0⁰.</p><p><strong>Step 2:</strong> Use logarithmic transformation. Let y = (sin x)^(tan x). Then:</p><p>ln y = tan x · ln(sin x)</p><p><strong>Step 3:</strong> Evaluate lim(x → 0⁺) ln y = lim(x → 0⁺) [tan x · ln(sin x)]. Rewrite as:</p><p>lim(x → 0⁺) ln y = lim(x → 0⁺) [ln(sin x)/(cot x)]</p><p>This is now in the form -∞/∞, suitable for L'Hôpital's rule.</p><p><strong>Step 4:</strong> Apply L'Hôpital's rule:</p><p>lim(x → 0⁺) [ln(sin x)/(cot x)] = lim(x → 0⁺) [(cos x/sin x)/(-csc² x)]</p><p>= lim(x → 0⁺) [(cos x/sin x) · (-sin² x)]</p><p>= lim(x → 0⁺) [-cos x · sin x]</p><p><strong>Step 5:</strong> Evaluate the limit:</p><p>lim(x → 0⁺) [-cos x · sin x] = -cos(0) · sin(0) = -1 · 0 = 0</p><p><strong>Step 6:</strong> Since lim(x → 0⁺) ln y = 0, we have:</p><p>lim(x → 0⁺) y = e⁰ = 1</p><p><strong>∴ Answer: 1</strong></p>
Correct Answer: 1