can u solve the given problem....
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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
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
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