limh→0ln(1+h)h=1 since:
(1) ln(a)−ln(b)=ln(ab) since:
- Let y=ln(a)−ln(b)
- ey=eln(a)−ln(b)=eln(a)eln(b)=ab
- y=ln(ab)
(2) limh→0ln(x+h)−ln(x)h=limh→0ln(1+hx)h
(3) Let t=hx so that h=xt and:
limh→0ln(1+hx)h=limt→0ln(1+t)xt=limt→0(1xt)ln(1+t)
(4) nln(a)=ln(an) since:
- Let y=nln(a)
- ey=enln(a)=(eln(a))n=an
- y=ln(an)
(5) limt→0(1xt)ln(1+t)=limt→0ln([1+t]1xt)=limt→0ln([(1+t)1t]1x)=(1x)limt→0ln([1+t]1t)
(1x)limt→0ln([1+t]1t)=(1x)limu→∞ln([1+1u]u)=(1x)ln(e)=1x
(7) Since limh→0ln(x+h)−ln(x)h=1x, it follows that:
limh→0ln(1+h)h=limh→0ln(1+h)−ln(1)h=11=1
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