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 M 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(2x) = 2 sin(x) cos(x)
Also, in the numerator we have
cos⁴(x) - cos²(x) = cos²(x) (cos²(x) - 1) = - cos²(x) sin²(x) = -1/4 sin²(2x)
So we can simplify M 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 x and sin(nx) all go to 0, and we're left with
Then the original limit is