Question

`lim_(x->0)(x/sinx)^(3/x^2)=` ?

Answer 1

Rewrite and use l'Hopital's Rule multiple times.

Explanation:

The initial form is indeterminate `1^oo`.

A common way of dealing with this form is to rewrite as `e` to a power, then find the limit of the exponent. By continuity of `e^x`, this lets us find the limit.

`(x/sinx)^(3/x^2) = e^(3/x^2ln(x/sinx))`

We want to find

`lim_(xrarr0)(3ln(x/sinx))/x^2 = 3lim_(xrarr0)(ln(x/sinx))/x^2`

This limit has form `0/0` so we use l'Hopital:

`= 3lim_(xrarr0)(sinx/x ((sinx-xcosx)/(sin^2x)))/(2x)`

`= 3/2 lim_(xrarr0)((sinx-xcosx)/(x^2sinx))`

This limit also has form `0/0`, so apply l'Hopital again.

`= 3/2 lim_(xrarr0)((cosx-(cosx-xsinx))/(2xsinx + x^2cosx))`

`= 3/2 lim_(xrarr0)((sinx)/(2sinx + xcosx))`

Oh, No! It's STILL `0/0`!

Yes, but the derivative of the numerator is `cosx`. So one more application of l'Hopital should get us an answer.

`= 3/2 lim_(xrarr0)((cosx)/(2cosx + (cosx-xsinx)))`

`= 3/2(1/(2+1+0)) = 3/2 (1/3) = 1/2`

Therefore,

`lim_(xrarr0)(x/sinx)^(3/x^2) = lim_(xrarr0) e^(3/x^2ln(x/sinx))`

`= e^(lim_(xrarr0) (3/x^2ln(x/sinx)))`

`= e^(1/2) = sqrte`

Answer 2

`sqrt(e)`

Explanation:

`sin x = x-x^3/(3!)+x^5/(5!)+ cdots` and
`sin x/x = 1-x^2/(3!)+x^4/(5!)+ cdots`

both series are alternate and for `abs x le 1` we have

`1-x^2/(3!) le sin x/x le 1-x^2/(3!)+x^4/(5!)` and also

`1/(1-x^2/(3!)+x^4/(5!)) le x/sin x le 1/(1-x^2/(3!) )` and also

`(1/(1-x^2/(3!)+x^4/(5!)))^(3/x^2) le ( x/sin x)^(3/x^2) le (1/(1-x^2/(3!) ))^(3/x^2)`

but

`lim_(x->0) (1/(1-x^2/(3!) ))^(3/x^2) = lim_(x->0) (1/(1-x^2/(3!) ))^((3!)/x^2/2)` and making `y = x^2/(3!)` we have

`lim_(x->0) (1/(1-x^2/(3!) ))^((3!)/x^2/2) =sqrt(lim_(y->0)(1/(1-y))^(1/y)) = sqrt(e)`

Analogously for

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

so

`lim_(x->0)(x/sinx)^(3/x^2)= sqrt(e)`