evaluate the limit:

lim x----> 0  ( sinx  /  x )^(1/x^2)

answer is given as e^-1/6 .

plz tell the method or any clue, rates assured.

 lim x--> 0 (sinx/x)^(1/x²) = [1+(sinx-x)/x]^(1/x²)

Now the above limit is of the type (1 + f)^g , where f---->0 and g----> infinity

Hence, the result is e^f*g  = lim x----> 0 e^{(sinx-x)/x^3} = e^-1/6 (By using expansions)