用泰勒公式证明极限题目!lim(x→0)=(e的x次方-sinx-1)/[1-根号下(1-x*x)]
问题描述:
用泰勒公式证明极限题目!
lim(x→0)=(e的x次方-sinx-1)/[1-根号下(1-x*x)]
答
没有分谁干呀?
答
=lim[1+x+(1/2)x^2+(1/6)x^3+……-x+(1/6)x^3+……-1]/[1-1+(1/2)x^2+……]
=lim[(1/2)x^2+o(x^2)]/[(1/2)x^2+o(x^2)]
=lim(1/2+o(x^2)/x^2]/[1/2+o(x^2)/x^2]
=(1/2+0)/(1/2+0)
=1
答
e^x=1+x+x^2/2+o(x^2)
sinx=x+o(x^2)
所有,e^x-sinx-1=1/2×x^2+o(x^2)
√(1-x^2)=1-1/2×x^2+o(x^2),所以1-√(1-x^2)=1/2×x^2+o(x^2)
lim(x→0) [e^x-sinx-1]/[1-√(1-x^2)]
=lim(x→0) [1/2×x^2+o(x^2)]/[1/2×x^2+o(x^2)]
=lim(x→0) [1/2+o(x^2)/x^2]/[1/2+o(x^2)/x^2]
=[1/2+0]/[1/2+0]=1