这是 1^∞ 型,先计算 lim(x→∞)x*lncos(1/x) (∞*0) = lim(t→0)lncost/t (0/0) = lim(x→0)(sint/cost)/1 = 0,所以 lim(x→0+)[cos(1/x)]^x = e^lim(x→∞)x*lncos(1/x) = e^0 = 1....
这题为什么不能用洛必达法则连续求导
这是 1^∞ 型,先计算 lim(x→∞)x*lncos(1/x) (∞*0) = lim(t→0)lncost/t (0/0) = lim(x→0)(sint/cost)/1 = 0,所以 lim(x→0+)[cos(1/x)]^x = e^lim(x→∞)x*lncos(1/x) = e^0 = 1....