Question

高数 求具体步骤

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Answer 1

1) 原式=lim(x→1)[lnx-(x-1)]/[(x-1)lnx]
=lim(x→1)(1/x-1)/[lnx+(x-1)/x]
=lim(x→1)[(1-x)/x]/[(xlnx+x-1)/x]
=lim(x→1)(1-x)/(xlnx+x-1)
=lim(x→1)-1/(lnx+x/x+1)
=-1/(ln1+1+1)
=-1/2
2) sin2x~2x，sinx~x
原式=2x/x
=2
3) sinx~x
原式=lim(x→0)(1-sinx/x)
=1-x/x
=1-1
=0
4) 原式=lim(x→∞)[1-3/x^2+2/x^3]/[1-1/x-1/x^2+1/x^3]
=1/1
=1
5) 原式=lim(x→0)[xsin(1/x)]/[sinx/x]
=0/[x/x]
=0/1
=0
6) lim(x→0-)f(x)
=lim(x→0-)(1-x)
=1-0
=1
lim(x→0+)f(x)
=lim(x→0+)(x^2+1)
=0+1
=1
lim(x→0-)f(x)=lim(x→0+)f(x)=1
∴lim(x→0)f(x)=1

Answer 2

追答

Answer 3

2.3.4都可以用洛必达法则吧，第六个分别求左右极限