# What is limit (sin x)^(1/2x-pie) if x --- pie/2?

sciencesolve | Certified Educator

You need to evaluate the limit, hence, you should replace `pi/2` for `x` such that:

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

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

`lim_(x->pi/2) (sin x)^(1/(2x-pi)) = 1^(1/0)`

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

The indetermination `1^oo` requests for you to use the following approach, such that:

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

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

Using the power property of logarithms yields:

`lim_(x->pi/2) ln (sin x)^(1/(2x-pi)) = lim_(x->pi/2) (ln (sin x))/(2x-pi) = (ln 1)/(pi-pi) = 0/0`

The indetermination `0/0` requests for you to use l'Hospital's theorem, such that:

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

`lim_(x->pi/2) ((ln (sin x))')/((2x-pi)') = lim_(x->pi/2) (cos x/sin x)/2 = 0/2 = 0`

Replacing 0 for `lim_(x->pi/2) ln (sin x)^(1/(2x-pi))` yields:

`lim_(x->pi/2) e^(ln (sin x)^(1/(2x-pi))) = e^0 = 1`

Hence, evaluating the given limit, using the indicated approach and l'Hospital's theorem, yields `lim_(x->pi/2) (sin x)^(1/(2x-pi)) = 1.`