f(x)=(tanx)^2((2(sinx)^2+ 3sinx+4))^0.5 - ((sinx)^2 +6sinx +2)^0.5

f Defined from o to pi closed intervals to all R

Find limit at x=pi/2

The given function can be rewritten as
F(x) = tan^2 x [ 2 sin^2 x+ 3sin x +4] –[sin^2x +6sinx +2)^1/2
When x approaches pi/2 sinx =1 So this becomes
F(pi/2) = tan^2 x ( 2+3+4)^1/2 – (1+6+2)^1/2 = tan^2 x (3) -3
As x tends to pi/2 tan x tends to infinity.
So required limit is + infinity.
Regards
Nirmal Singh
Askiitians Faculty