1,o/o型,用洛必達(dá)法則原式=lim(x→1) cosx/1=cos12,∵√(1+cosx)=√{2[cos(x/2)]^2}=√2*cos(x/2)∴√2-√(1+cosx)=√2*[1-cos(x/2)]等價(jià)無(wú)窮?。寒?dāng)x→0時(shí),cosx=1-1/2*x^2,sinx=2x∴原式=lim(x→0) √2*[1-cos(x/2)]...第二題我打錯(cuò)了，分母是sin^2x∵√(1+cosx)=√{2[cos(x/2)]^2}=√2*cos(x/2)
∴√2-√(1+cosx)=√2*[1-cos(x/2)]
等價(jià)無(wú)窮?。寒?dāng)x→0時(shí)，cosx=1-1/2*x^2，sinx=x，
∴cos(x/2)=1-1/2*(1/2*x)^2=1-1/8*x^2
∴原式=lim(x→0) √2*[1-cos(x/2)]/(sinx)^2
=lim(x→0) √2*1/8*x^2/x^2
=√2/8