Original formula = lim (x→ 0) e [(cotx 2) lncosx]

=lim(x→0) e^[lncosx/tanx^2]

= lim (x→ 0) e [lncosx/x 2] (infinitesimal equivalent substitution)

= lim (x → 0) e [(-sinx/cosx)/2x] (Lobida rule, use as little as possible)

When x→0, 1/cosx= 1, sinx ~ x.

So the original formula = = lim (x→ 0) e (- 1/2).