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