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