这是个1＾∞极限,先取自然对数
lim(x→∞）ln(cos1/x)^x^2
=lim(x→∞）x^2ln(cos1/x) (1/x=t,t→0)
=lim(t→0）ln(cost) /t^2 (运用洛必达法则）
=lim(t→0）(-sint/cost) /(2t)
=lim(t→0）-sint /(2t)=-1/2
所以
lim(x→∞）(cos1/x)^x^2=lim(x→∞）e^ln(cos1/x)^x^2=e^(-1/2)