Solution
Related Formula
Standard Taylor expansions near zero:
h = 1 - (h²)/(2!) + (h⁴)/(4!) - h = h - (h³)/(3!) + (h⁵)/(5!) -Core Logic
Let x - 1 = h, where h → 0⁺. The expression transforms into:
h → 0(h(6 + λ h) - μ h)/(h³) = -1Substitute the expansions into the numerator:
h → 0(h[6 + λ(1 - (h²)/(2))] - μ(h - (h³)/(6)))/(h³) = -1 h → 0((6 + λ - μ)h + (-(λ)/(2) + (μ)/(6))h³)/(h³) = -1Step 1: Match Coefficients for Existence
For the limit to be finite, the coefficient of h must vanish:
6 + λ - μ = 0 μ - λ = 6 (1)Equating the h³ term to the given limit value:
-(λ)/(2) + (μ)/(6) = -1 -3λ + μ = -6 (2)Step 2: Solve System of Equations
Subtract equation (1) from (2):
(-3λ + μ) - (μ - λ) = -6 - 6 -2λ = -12 λ = 6From (1), μ = 6 + 6 = 12.
λ + μ = 6 + 12 = 18Pattern Recognition
When dealing with indeterminate form limits involving mixed trigonometric expressions with a non-zero denominator power, polynomial substitution using Taylor series is much cleaner and less prone to differentiation tracking mistakes compared to multiple L'Hôpital cycles.
Chapter Mix
Class 12 Mathematics: Limits, Continuity and Differentiability