Numerator: loge(sec(ex)sec(e2x)⋯sec(e10x))=k=1∑10loge(sec(ekx)).
Using sec(u)≈1+2u2 for small u, we get loge(sec(ekx))≈2e2kx2.
Thus the numerator is ≈2x2k=1∑10e2k=2(e2−1)x2e2(e20−1).
Denominator: Using cosx≈1−2x2, we have e2cosx≈e2(1−x2), so e2−e2cosx≈e2x2.
Therefore: x→0lim=e2x22(e2−1)x2e2(e20−1)=2(e2−1)e20−1.