63 - Calculus Questions Answers
limit n tends to ∞
then
[³√(n²-n³) + n ] equals
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
if f(X)=xlxl then find f-1(x) can you please explain me the meaning and use of sgn
f(x) = x|x| => f(x) = -x2 for negative real values of x and f(x) = x2 for positive real values of x
so f-1(x) = -√|x| for negative real values and = +√|x| for positive real values of x
which of the following is differntiable at x=0 ?
cos(lxl)+lxl
sin(lxl)-lxl
sin(|x|)+|x| is differentiable at x = 0
LHD of first = {cos|0+h|+|0+h|-cos|0|-|0|}/0+h-0 = (cosh+h-1)/h = (1-2sin2h+h-1)/h = 1 (on taking limits)
now check RHD, its value will be -1
similarly in second both values are 0 so it is differentiable