Limits, Continuity & Differentiability
Limits of Inverse Trigonometric Functions
Grade 12
<p>Let \(f(x) = \cot^{-1}\left(\dfrac{x^{2018}+5}{(x-5)(x-10)}\right)\), then:</p>
<p>\(\lim_{x \to 5^-} f(x) = 0\)</p>
<p>\(\lim_{x \to 5^+} f(x) = \pi\)</p>
<p>\(\lim_{x \to 10^-} f(x) = \pi\)</p>
<p>\(\lim_{x \to 10^+} f(x) = 0\)</p>
Step-by-Step Solution
Key Concept: Analyze the domain and behavior of the argument of cot⁻¹ by examining the sign and magnitude of the rational expression (x²⁰¹⁸+5)/((x-5)(x-10)) across different intervals. The cot⁻¹ function has domain ℝ and outputs (0,π), with critical behavior determined by whether the argument approaches ±∞.
<p><strong>Step 1:</strong> Identify critical points where denominator = 0: x = 5 and x = 10. The numerator x²⁰¹⁸ + 5 > 0 for all x ∈ ℝ.</p><p><strong>Step 2:</strong> Analyze sign of denominator (x-5)(x-10):</p><ul><li>x < 5: both factors negative → denominator > 0</li><li>5 < x < 10: (x-5) > 0, (x-10) < 0 → denominator < 0</li><li>x > 10: both factors positive → denominator > 0</li></ul><p><strong>Step 3:</strong> Evaluate behavior at critical points:</p><ul><li>As x → 5⁻: argument → +∞ ⟹ f(5⁻) → cot⁻¹(+∞) = 0</li><li>As x → 5⁺: argument → -∞ ⟹ f(5⁺) → cot⁻¹(-∞) = π</li><li>As x → 10⁻: argument → -∞ ⟹ f(10⁻) → cot⁻¹(-∞) = π</li><li>As x → 10⁺: argument → +∞ ⟹ f(10⁺) → cot⁻¹(+∞) = 0</li></ul><p><strong>Step 4:</strong> Determine continuity and differentiability:</p><ul><li>At x = 5: left limit = 0, right limit = π → discontinuous (jump of π)</li><li>At x = 10: left limit = π, right limit = 0 → discontinuous (jump of π)</li><li>On intervals (-∞,5), (5,10), (10,∞): f is continuous and differentiable</li></ul><p>∴ Answer: f is continuous and differentiable on (-∞,5); f is continuous and differentiable on (5,10); f is continuous and differentiable on (10,∞); f is discontinuous at both x=5 and x=10</p>
Correct Answer: A,B,C,D