Question

can u solve the given problem....

Answer 1

Apply L Hospital's Rule,

Lt x->0       [sec²x e^(tanx) - cosx  e^sinx  ] / -4sin 4x

Applying again

Lt  x->0     
(  [sec^4 x e^tan x + (e^tanx ) 2 sec²x tan x]  -  [cos²x e^(sinx) +                                                                                                        e^sinx (-sinx ) ] )
divided by -16 sin 4x

Put x = 0

The answer should be -1/16