Question
limit n tends to ∞
then
[³√(n²-n³) + n ] equals
Joshi sir comment
We have to calculate limit n -> ∞ [{³√(n²-n³) }+ n ]
Use the formula a3+b3 = (a+b)(a2+b2-ab) in the format
(a+b) = (a3+b3)/(a2+b2-ab)
here a = ³√(n²-n³) and b = n
on solving we get 1/(1+1+1) = 1/3