2个回答
展开全部
f(x)
=(e^x-1)/x ; x≠0
=1 ; x=0
lim(x->0) f(x)
=lim(x->0) (e^x-1)/x
=1
=f(0)
x=0, f(x) 连续
f'(0)
=lim(h->0) [f(h) -f(0)]/h
=lim(h->0) [ (e^h-1)/h -1]/h
=lim(h->0) (e^h-1-h)/h^2
=lim(h->0) (1/2)h^2/h^2
=1/2
x->0
e^x = 1+x+(1/2)x^2+...
f(x) =(e^x-1)/x = 1+(1/2)x + (1/3!)x^2+....+(1/(n+1)!)x^n+....
f^(n)(x)
=(1/(n+1)!).n! + o(x^0)
=1/(n+1) +o(x^0)
f^(n)(0) = 1/(n+1)
=(e^x-1)/x ; x≠0
=1 ; x=0
lim(x->0) f(x)
=lim(x->0) (e^x-1)/x
=1
=f(0)
x=0, f(x) 连续
f'(0)
=lim(h->0) [f(h) -f(0)]/h
=lim(h->0) [ (e^h-1)/h -1]/h
=lim(h->0) (e^h-1-h)/h^2
=lim(h->0) (1/2)h^2/h^2
=1/2
x->0
e^x = 1+x+(1/2)x^2+...
f(x) =(e^x-1)/x = 1+(1/2)x + (1/3!)x^2+....+(1/(n+1)!)x^n+....
f^(n)(x)
=(1/(n+1)!).n! + o(x^0)
=1/(n+1) +o(x^0)
f^(n)(0) = 1/(n+1)
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展开全部
f'(x)=(e^x.x-(e^x-1))/x²
=(e^x.(x-1)+1)/x²
f'(0)=lim(x-->0)(e^x.(x-1)+1)/x²
=lim(x-->0)(e^x.(x-1)+e^x)/2x
=lim(x-->0)e^x/2
=1/2
或者根据导数定义:(洛必达)
f'(0)=lim(x-->0)[f(x)-f(0)]/x
=lim(x-->0)[(e^x-1)/x-1]/x
=lim(x-->0)[(e^x-1)-x]/x²
=lim(x-->0)[e^x-1]/2x
=lim(x-->0)[e^x]/2
=1/2
f(x)=x^(-1)e^x-x^(-1)
前面1项用牛顿-莱布尼兹公式:
(uv)^(n)=Σ(i=0,n)C(n,i)u^(i)v^(n-i)
后面一项:
(x^(-1))^(n)=(-1)(-2)...(-n)x^(-n-1)
=(-1)^n.n!x^[-(n+1)]
f^(n)(x)=Σ(i=0,n)C(n,i)(x^(-1))^(i)(e^x)^(n-i)-(-1)^n.n!x^[-(n+1)]
=(e^x)Σ(i=0,n)C(n,i)(-1)^i.i!x^[-(i+1)]-(-1)^n.n!x^[-(n+1)]
=(e^x)Σ(i=0,n)n!/i!/(n-i)!(-1)^i.i!x^[-(i+1)]-(-1)^n.n!x^[-(n+1)]
=(e^x)Σ(i=0,n)(-1)^i.n!/(n-i)!.(x^(-1)]^(i+1)-(-1)^n.n![x^(-1)]^(n+1)
设1/x=t,x=1/t
f^(n)(1/t)=e^(1/t).[t-nt²+n(n-1)t³-...+(-1)^n.n!.t^(n+1)-(-1)^n.n!.t^(n+1)
f''(x)=e^x.[1/x-2.1/x²+2.1/x³]-2.1/x³
=e^x.[x²-2x+2]/x³-2/x³
f''(0)=lim(x-->0){e^x.[x²-2x+2]/x³-2/x³}
=lim(x-->0){e^x.(x²-2x+2)-2]/x³
=lim(x-->0){e^x.(x²-2x+2+2x-2)]/3x²
=lim(x-->0){e^x]/3
=1/3
推论:
f^(n)(0)=1/(n+1)
f'''(x)={e^x.[x²-2x+2+2x-2]x³-e^x.[x²-2x+2].3x²}/x^6+6/x^4
={e^x.x³-e^x.[x²-2x+2].3}/x^4+6/x^4
=e^x.(x³-3x²+6x-6)/x^4+6/x^4
=[e^x.(x³-3x²+6x-6)+6]/x^4
f'''(0)=lim(x-->0)[e^x.(x³-3x²+6x-6)+6]/x^4
=lim(x-->0)[e^x.(x³-3x²+6x-6+3x²-6x+6)]/4x³
=lim(x-->0)[e^x.(x³)]/4x³
=lim(x-->0)e^x./4
=1/4
=(e^x.(x-1)+1)/x²
f'(0)=lim(x-->0)(e^x.(x-1)+1)/x²
=lim(x-->0)(e^x.(x-1)+e^x)/2x
=lim(x-->0)e^x/2
=1/2
或者根据导数定义:(洛必达)
f'(0)=lim(x-->0)[f(x)-f(0)]/x
=lim(x-->0)[(e^x-1)/x-1]/x
=lim(x-->0)[(e^x-1)-x]/x²
=lim(x-->0)[e^x-1]/2x
=lim(x-->0)[e^x]/2
=1/2
f(x)=x^(-1)e^x-x^(-1)
前面1项用牛顿-莱布尼兹公式:
(uv)^(n)=Σ(i=0,n)C(n,i)u^(i)v^(n-i)
后面一项:
(x^(-1))^(n)=(-1)(-2)...(-n)x^(-n-1)
=(-1)^n.n!x^[-(n+1)]
f^(n)(x)=Σ(i=0,n)C(n,i)(x^(-1))^(i)(e^x)^(n-i)-(-1)^n.n!x^[-(n+1)]
=(e^x)Σ(i=0,n)C(n,i)(-1)^i.i!x^[-(i+1)]-(-1)^n.n!x^[-(n+1)]
=(e^x)Σ(i=0,n)n!/i!/(n-i)!(-1)^i.i!x^[-(i+1)]-(-1)^n.n!x^[-(n+1)]
=(e^x)Σ(i=0,n)(-1)^i.n!/(n-i)!.(x^(-1)]^(i+1)-(-1)^n.n![x^(-1)]^(n+1)
设1/x=t,x=1/t
f^(n)(1/t)=e^(1/t).[t-nt²+n(n-1)t³-...+(-1)^n.n!.t^(n+1)-(-1)^n.n!.t^(n+1)
f''(x)=e^x.[1/x-2.1/x²+2.1/x³]-2.1/x³
=e^x.[x²-2x+2]/x³-2/x³
f''(0)=lim(x-->0){e^x.[x²-2x+2]/x³-2/x³}
=lim(x-->0){e^x.(x²-2x+2)-2]/x³
=lim(x-->0){e^x.(x²-2x+2+2x-2)]/3x²
=lim(x-->0){e^x]/3
=1/3
推论:
f^(n)(0)=1/(n+1)
f'''(x)={e^x.[x²-2x+2+2x-2]x³-e^x.[x²-2x+2].3x²}/x^6+6/x^4
={e^x.x³-e^x.[x²-2x+2].3}/x^4+6/x^4
=e^x.(x³-3x²+6x-6)/x^4+6/x^4
=[e^x.(x³-3x²+6x-6)+6]/x^4
f'''(0)=lim(x-->0)[e^x.(x³-3x²+6x-6)+6]/x^4
=lim(x-->0)[e^x.(x³-3x²+6x-6+3x²-6x+6)]/4x³
=lim(x-->0)[e^x.(x³)]/4x³
=lim(x-->0)e^x./4
=1/4
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