AP EAMCET201921 Apr 2019Evening ShiftMathematicsLimitsActual
If α = lim x → 0 x · 2 x - x 1 - cos x and β = lim x → 0 x · 2 x - x 1 + x 2 - 1 - x 2 , then
Options
- Aα = 5 β
- Bα = 2 β
- Cβ = 2 α 2
- Dβ = 1 6 α
Correct answer
B. α = 2 β
Step-by-step solution
It is given that, α = lim x → 0 x · 2 x - x 1 - cos x Apply L hospital rule, a = lim x → 0 2 x + x · 2 x log 2 - 1 sin x Again, apply L hospital rule, α = lim x → 0 2 x log 2 + 2 x log 2 + x · 2 x ( log 2 ) 2 cos x α = log 2 + log 2 α = 2 log 2   … i Now β = lim x → 0 x · 2 x - x 1 + x 2 - 1 - x 2 Apply L hospital rule, β = lim x → 0 2 x + x · 2 x log 2 - 1 x 1 + x 2 + x 1 - x 2 Again, apply L hospital rule, β = lim