Question

How do you find the limit of `x^(sin(x))` as x approaches 0?

Answer 1

`1`

Explanation:

let `L = lim_(x to 0) x^(sin x)`

`implies ln L = ln lim_(x to 0) x^(sin x)`

`= lim_(x to 0) ln x^(sin x)`

`= lim_(x to 0) sinx ln x`

`= lim_(x to 0) (ln x)/(1/(sinx) )`

`= lim_(x to 0) (ln x)/(csc x )`

this is in indeterminate `oo/oo` form so we can use L'Hôpital's Rule

`= lim_(x to 0) (1/x)/(- csc x cot x)`

`=- lim_(x to 0) (sin x tan x)/(x)`

Next bit is unnecessary, see ratnaker-m's note below...

this is now in indeterminate `0/0` form so we can go again

`ln L =- lim_(x to 0) (cos x tan x + sin x sec^2 x)/(1)`

`= - 0`

So:
`L = e^(- 0) = 1`

Answer 2

`1`

Explanation:

`x^sin x = (1+x-1)^sinx = 1 +sin x/(1!)(x-1)+(sinx(sinx-1))/(2!)(x-1)^2+cdots+`

then

`lim_(x->0)x^sinx = 1`