Limits, Continuity & Differentiability
Evaluation of Trigonometric Limits
Grade 12
Question:
<p>Find the value of \[\lim_{x \to 0} \frac{(1-\cos 2x)(3+\cos x)}{x\tan 4x}\]</p>
<p>1</p>
<p>2</p>
<p>3</p>
<p>4</p>
Step-by-Step Solution
Key Concept: Use the Taylor series approximations cos(2x) ≈ 1 - 2x² and tan(4x) ≈ 4x near x = 0, combined with recognizing that (1 - cos 2x) = 2sin²(x) to simplify the numerator and denominator efficiently.
<p><strong>Step 1:</strong> Identify the indeterminate form. As x → 0: numerator → (1-1)(3+1) = 0 and denominator → 0·0 = 0, so we have 0/0 form.</p><p><strong>Step 2:</strong> Use key approximations near x = 0:</p><ul><li>1 - cos(2x) = 2sin²(x) ≈ 2x²</li><li>3 + cos(x) → 3 + 1 = 4</li><li>tan(4x) ≈ 4x</li></ul><p><strong>Step 3:</strong> Substitute approximations:</p><p>$$\lim_{x \to 0} \frac{(1-\cos 2x)(3+\cos x)}{x\tan 4x} = \lim_{x \to 0} \frac{2x^2 \cdot 4}{x \cdot 4x}$$</p><p><strong>Step 4:</strong> Simplify:</p><p>$$= \lim_{x \to 0} \frac{8x^2}{4x^2} = \frac{8}{4} = 2$$</p><p>∴ Answer: <strong>B (which equals 2)</strong></p>
Correct Answer: B