You need to evaluate the following limit such that:
`lim_(x->0) (sin^2 x)/(1-2sin((5pi)/6 - x)cos x) = (sin^2 0)/(1 - 2sin((5pi)/6)*cos 0)`
Since `sin 0 = 0 ` and `cos 0 = 1` yields:
`lim_(x->0) (sin^2 x)/(1-2sin((5pi)/6 - x)cos x) = 0/(1 - 2sin((5pi)/6))`
You need to evaluate `sin((5pi)/6) = sin(pi/3 + pi/2)`
Using the identity `sin(a+b) = sin a*cos b + sin b*cos a` yields:
`sin(pi/3 + pi/2) = sin(pi/3)cos(pi/2) + sin(pi/2)cos(pi/3)`
Since `sin(pi/2) = 1` and `cos(pi/2) = 0` yields:
`sin(pi/3 + pi/2) = cos(pi/3) = 1/2`
`lim_(x->0) (sin^2 x)/(1-2sin((5pi)/6 - x)cos x) = 0/(1 - 2(1/2))`
`lim_(x->0) (sin^2 x)/(1-2sin((5pi)/6 - x)cos x) = 0/(1 - 1) = 0/0`
Since the limit is indeterminate 0/0, you may use l'Hospital's theorem such that:
`lim_(x->0) (sin^2 x)/(1-2sin((5pi)/6 - x)cos x) = lim_(x->0) ((sin^2 x)')/((1-2sin((5pi)/6 - x)cos x)')`
`lim_(x->0) (sin^2 x)/(1-2sin((5pi)/6 - x)cos x) = lim_(x->0) (2 sin x cos x)/(-2cos((5pi)/6 - x)*(-1)*cos x + 2sin((5pi)/6 - x)sin x)`
`(-2/2)lim_(x->0) (sin x cos x)/(cos((5pi)/6 - x)*(-1)*cos x- sin((5pi)/6 - x)sin x) = -lim_(x->0) (sin x cos x)/(cos((5pi)/6 - x + x))`
`-lim_(x->0) (sin x cos x)/(cos((5pi)/6)) = -1/(cos((5pi)/6)) lim_(x->0) (sin x cos x)`
`-1/(cos((5pi)/6)) lim_(x->0) (sin x cos x) = -1/(cos((5pi)/6))sin 0*cos 0`
`-1/(cos((5pi)/6)) lim_(x->0) (sin x cos x) = -1/(cos((5pi)/6))*0*1 = 0`
Hence, evaluating the given limit, using l'Hospital's theorem, yields `lim_(x->0) (sin^2 x)/(1-2sin((5pi)/6 - x)cos x) = 0` .
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