Limits, Continuity & Differentiability
Trigonometric and Inverse Trigonometric Limits
Grade 12
Question:
<p>If $$I_1 = \lim_{x \to \infty} (\tan^{-1}(-x) - \tan^{-1}x)\cos x$$ and $$I_2 = \lim_{x \to 0} (\tan^{-1}(-x) - \tan^{-1}x)\cos x$$, then $$(I_1, I_2)$$ is</p>
<p>(a) $$(0, 0)$$</p>
<p>(b) $$(0, 1)$$</p>
<p>(c) $$(1, 0)$$</p>
<p>(d) None of the above</p>
Step-by-Step Solution
Key Concept: Use the property that tan⁻¹(-x) = -tan⁻¹(x) to simplify the difference, then evaluate the limits by analyzing the behavior of the expression as x approaches different values.
<p><strong>Step 1: Simplify the expression inside the limits</strong></p><p>Using the property of inverse tangent: tan⁻¹(-x) = -tan⁻¹(x)</p><p>tan⁻¹(-x) - tan⁻¹(x) = -tan⁻¹(x) - tan⁻¹(x) = -2tan⁻¹(x)</p><p></p><p><strong>Step 2: Rewrite I₁</strong></p><p>I₁ = lim(x→∞) [-2tan⁻¹(x)]·cos(x)</p><p>I₁ = -2·lim(x→∞) [tan⁻¹(x)·cos(x)]</p><p>As x→∞: tan⁻¹(x) → π/2 (a finite value)</p><p>As x→∞: cos(x) oscillates between -1 and 1 (does not converge)</p><p>The product of (finite value) × (oscillating function) does not have a limit. The limit does not exist.</p><p></p><p><strong>Step 3: Rewrite I₂</strong></p><p>I₂ = lim(x→0) [-2tan⁻¹(x)]·cos(x)</p><p>As x→0: tan⁻¹(x) → 0</p><p>As x→0: cos(x) → 1</p><p>I₂ = -2·(0)·(1) = 0</p><p></p><p><strong>Step 4: Analyze I₁ more carefully</strong></p><p>Since I₁ involves lim(x→∞) of a product where tan⁻¹(x)→π/2 but cos(x) oscillates, the limit does not exist in the classical sense. This is not among the given pairs (0,0), (0,1), or (1,0).</p><p></p><p>∴ Answer: Not specified</p>
Correct Answer: Not specified