Problem in solving an indeterminate limit [0/0] `lim_(x->0) (sin^2 x)/(1-2sin((5pi)/6 - x)cos x) ` Because I find it easier to solve it if I switch...

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` ...