If f x = cos x x 1 2 sin x x 2 2 x tan x x 1 , then lim x → 0 ⁡ f ′ x x

If fx=cosxx12sinxx22xtanxx1, then limx→0⁡f′xx
  1. does not exist
  2. exists and is equal to -2
  3. exists and is equal to 0
  4. exists and is equal to 2

Solution

Given fx=cosxx12sinxx22xtanxx1

=cosxx2-2x2-2sinxx-x+tanx2x2-x2

=-x2cosx+tanx.x2

=x2tanx-cosx

⇒ Limx→0f'xx=Limx→02xtanx-cosx+x2sec2x+sinxx

=Limx→02tanx-cosx+xsec2x+sinx

=-2

Asked in: JEE Main 2018 (15 Apr)

Practice more Determinants questions on Aicharya