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

##### 2 Answers

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

#### Explanation:

The initial form is indeterminate

A common way of dealing with this form is to rewrite as

We want to find

This limit has form

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

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

This limit also has form

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

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

Oh, No! It's STILL

Yes, but the derivative of the numerator is

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

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

Therefore,

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

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

#### Explanation:

both series are alternate and for

but

Analogously for

so