x->0
lim(cosx)^(1/x^2)
=lime^(lncosx)/x^2
=e^lim(lncosx)/x^2
x->0 l hospital法则
lim (lncosx)/x^2
=lim-sinx/2xcosx
=lim -1/2cosx
=-1/2
所以原式=e^(-1/2)