It looks like the limit you want to compute is

Rewrite the limand with an exponential and logarithm:

Now, since the exponential function is continuous at 0, we can write

Let <em>M</em> denote the remaining limit.
We have
as
, so
and
. Apply L'Hopital's rule:

Simplify and rewrite this in terms of sin and cos :

As
, we get another 0/0 indeterminate form. Apply L'Hopital's rule again:

Recall the double angle identity for sin:
sin(2<em>x</em>) = 2 sin(<em>x</em>) cos(<em>x</em>)
Also, in the numerator we have
cos⁴(<em>x</em>) - cos²(<em>x</em>) = cos²(<em>x</em>) (cos²(<em>x</em>) - 1) = - cos²(<em>x</em>) sin²(<em>x</em>) = -1/4 sin²(2<em>x</em>)
So we can simplify <em>M</em> as

This again yields 0/0. Apply L'Hopital's rule again:

Once again, this gives 0/0. Apply L'Hopital's rule one last time:

Now as
, the terms containing <em>x</em> and sin(<em>nx</em>) all go to 0, and we're left with

Then the original limit is
![L = \exp(M) = e^{-1/3} = \boxed{\dfrac1{\sqrt[3]{e}}}](https://tex.z-dn.net/?f=L%20%3D%20%5Cexp%28M%29%20%3D%20e%5E%7B-1%2F3%7D%20%3D%20%5Cboxed%7B%5Cdfrac1%7B%5Csqrt%5B3%5D%7Be%7D%7D%7D)