Question

Evaluate the limit `lim_(x→0)(sin x/x)^(1/x^2)`?

Answer 1

`e^(-1/6)`

Explanation:

`sinx = x-x^3/(3!)+x^5/(5!)+x^7/(7!)+ cdots`

this is an alternate series and if `abs x < 1`

`(x-x^3/(3!))/x < sin x/x < (x-x^3/(3!)+x^5/(5!))/x`

`lim_(x->0)((x-x^3/(3!))/x)^(1/x^2)=lim_(x->0)(1-x^2/6)^(1/x^2)`

but making `y = -x^2/6`

`lim_(x->0)(1-x^2/6)^(1/x^2) =lim_(y->0)((1+y)^(1/y))^(-1/6)=e^(-1/6)`

analogously

`lim_(x->0)( (x-x^3/(3!)+x^5/(5!))/x)^(1/x^2) = lim_(x->0) (1-x^2/(3!)+x^4/(5!))^(1/x^2) = e^(-1/6)`