<p>Let \(x_0 = \tan^{-1}(2)\) and \[b = \lim_{x \to \tan^{-1}2} \frac{(\tan^2 x - a)(1 + \tan x)}{e^{(\tan x - 2)} - 1}\] For the existence of the limit, find \([a + b + x_0]\) where \([\cdot]\) denotes the greatest integer function.</p>
Step-by-Step Solution
Key Concept: At x = tan⁻¹(2), tan x = 2 gives 0/0 form. For limit existence, numerator must also vanish at tan x = 2, requiring a = 4. Then apply L'Hôpital's rule or Taylor expansion around tan x = 2.
<p><strong>Step 1: Identify the indeterminate form</strong></p><p>When x → tan⁻¹(2), we have tan x → 2. At this point:</p><p>Denominator: e^(2-2) - 1 = e⁰ - 1 = 0</p><p>For limit to exist, numerator must also be 0:</p><p>(tan²x - a)(1 + tan x) → (4 - a)(1 + 2) = 3(4 - a) = 0</p><p>Therefore: <strong>a = 4</strong></p><p><strong>Step 2: Apply L'Hôpital's Rule</strong></p><p>With a = 4, the limit becomes 0/0 form.</p><p>b = lim(x→tan⁻¹2) [(tan²x - 4)(1 + tan x)] / [e^(tan x - 2) - 1]</p><p>Using L'Hôpital's rule (differentiate with respect to tan x since both numerator and denominator depend on tan x):</p><p><strong>Step 3: Compute derivative</strong></p><p>Numerator derivative: d/d(tan x)[(tan²x - 4)(1 + tan x)] = 2tan x(1 + tan x) + (tan²x - 4) = 3tan²x + 2tan x - 4</p><p>At tan x = 2: 3(4) + 2(2) - 4 = 12 + 4 - 4 = 12</p><p>Denominator derivative: d/d(tan x)[e^(tan x - 2) - 1] = e^(tan x - 2)</p><p>At tan x = 2: e⁰ = 1</p><p>Therefore: <strong>b = 12/1 = 12</strong></p><p><strong>Step 4: Calculate x₀</strong></p><p>x₀ = tan⁻¹(2) ≈ 1.1071 radians</p><p><strong>Step 5: Final Answer</strong></p><p>a + b + x₀ = 4 + 12 + 1.1071 = 17.1071</p><p>∴ [a + b + x₀] = [17.1071] = <strong>17</strong></p>
Correct Answer: 17