可根據(jù)導(dǎo)數(shù)的定義得
x趨于0,[1-cos(x^2)]/(1-cosx)的極限
=[cos0^2-cos(x^2)]/(cos0-cosx)的極限
=x{[cos0^2-cos(x^2)]/(0^2-x^2)}除以
[(cos0-cosx)/(0-x)] 的極限
=x乘以cos(x^2)的導(dǎo)數(shù)/cosx的數(shù) 的極限
=x*[-2sin(x^2)}/(-sinx) 的極限
=2sin(x^2)*[x/sinx] 的極限
=2sin(x^2)*[(x-0)/(sinx-sin0)] 的極限
=2sin(x^2)*[1/(sinx的導(dǎo)數(shù))] 的極限
=2sin(x^2)*(1/cosx) 的極限
=2sin(0^2)*(1/cos0)=0