# 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 numerator and denominator: I get that...

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 numerator and denominator: I get that lim(x->0)(1/f(x))=inf which means (probably) that 1/y=inf and then y=0.

### 1 Answer | Add Yours

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` .**