lim (sinx) ^ 1/x ; x is tending to 0 , x>0. how to solve this limit. plz give the full soluntion

its simple sin x/x ,,,series of sinx is  x-[(x^3)/6]  taking x common u get 1-[(x^2)/2] and now x tends to 0 so x^2 is negligible   

and ans is 1 or 

       use graph at x→0  (nearly zero) y=x and y= sinx takes same values as y=x is tangent to y=sinx at x=0