Question

`lim_(x->pi/2)(sinx)^(1/sin(2x)) =`?

Answer 1

undefined

Explanation:

Solve by direct substitution

`lim_(x->pi/2)(sin x)^(1/(sin 2x)) = (sin (pi/2))^(1/(sin (pi))`

`= 0^(1/0) = oo or "undefined"`

Apply L'Hopital's rule but to do this we have to modify the function

`y(x) = (sin x)^(1/(sin 2x))`

Take natural logarithm of both sides

`ln(y) = ln(sin x)^(1/(sin 2x)) = 1/(sin(x))ln(sin(2x))`

`ln(y) = ln(sin(x))/(sin(x))`

`lim_(x->pi/2)(sin x)^(1/(sin 2x)) = ln(sin(x))/(sin(2x))`

`= "undefined"`

Now we can apply L'Hopital's rule

Numerator

`d/dx sin(x) = cos(x)`

Denominator

`d/dx(sin(2x)) = d/(du)(sin(u))d/dx(2x)`

`= cos(u)*2`

`= cos(2x)*2`

Divide numerator by denominator

`cos(x)/(cos(2x)*2) = cos((pi)/2)/(cos(pi)*2) = 0/0`

So again we derivate numerator and denominator

Numerator

`d/dx cos(x) = -sin(x) = -1`

Denominator

`d/dx(cos(2x)*2) = 2d/(du)cos(u)d/dx(2x)`

`= -4sin(2x)`

`=0`

I used a calculator to find where the function becomes defined but you no matter how much you derivate the answer remain undefined

Numerator

`d/dx(-sin(x)) rarr -cos(x) rarr d/dx(-cos(x)) rarr sin(x) rarr sin(pi/2) = 1`

Denominator

`d/dx(-4sin(2x)) rarr -8\cos (2x\ ) rarr d/dx (-8\cos (2x\ )) rarr 16\sin \(2x\) rarr 16\sin \(pi\) = 0`

Answer 2

`lim_(x->pi/2)(sinx)^(1/sin(2x)) = 1`

Explanation:

Near `x = pi/2` we have

`sinx = 1-1/2(x-pi/2)^2+1/24(x-pi/2)^2+O((x-pi/2)^5)`

and

`sin(2x) = -2(x-pi/2)+4/3(x-pi/2)^3+O((x-pi/2)^5)`

Then making `x-pi/2 = y`

`lim_(x->pi/2)(sinx)^(1/sin(2x)) equiv lim_(y->0)(1-1/2y^2+O(y^4))^(1/(-2y+O(y^3)))`

`= lim_(y->0)1^(-1/(2y)) = 1`

`lim_(x->pi/2)(sinx)^(1/sin(2x)) = 1`