<p>Evaluate: \(\displaystyle\lim_{x \to \frac{\pi}{4}} \dfrac{\cot^3 x - \tan x}{\cos\left(x + \dfrac{\pi}{4}\right)}\)</p>
Step-by-Step Solution
Key Concept: Rewrite cot³x - tan x in terms of a common trigonometric function, then use the cosine addition formula to recognize that the denominator equals √2·sin(x - π/4), creating a telescoping structure with the numerator.
<p><strong>Step 1:</strong> Recognize 0/0 form at x = π/4. Rewrite the numerator:</p><p>cot³x - tan x = (cos³x/sin³x) - (sin x/cos x) = (cos⁴x - sin⁴x)/(sin³x cos x)</p><p><strong>Step 2:</strong> Factor cos⁴x - sin⁴x = (cos²x - sin²x)(cos²x + sin²x) = cos 2x · 1</p><p>So numerator = cos 2x/(sin³x cos x)</p><p><strong>Step 3:</strong> Rewrite denominator using cos(x + π/4) = (1/√2)(cos x - sin x)</p><p><strong>Step 4:</strong> Since cos 2x = (cos x - sin x)(cos x + sin x), the limit becomes:</p><p>lim[x→π/4] [(cos x - sin x)(cos x + sin x)]/[sin³x cos x] · √2/(cos x - sin x)</p><p><strong>Step 5:</strong> Cancel (cos x - sin x) terms (valid as x → π/4, not equal at limit point):</p><p>= lim[x→π/4] √2(cos x + sin x)/(sin³x cos x)</p><p><strong>Step 6:</strong> Substitute x = π/4: cos(π/4) = sin(π/4) = 1/√2</p><p>= √2 · (1/√2 + 1/√2)/[(1/√2)³ · (1/√2)] = √2 · (2/√2)/(1/4) = 2 · 4 = <strong>8</strong></p>
Correct Answer: C