2个回答
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答:
1)
lim(x→0) [1/x-1/(e^x-1)]
=lim(x→0) (e^x-1-x)/[x(e^x-1)]
=lim(x→0) (e^x-1-x) /x^2 (0----0型可导应用洛必达法则)
=lim(x→0) (e^x-1)/(2x)
=lim(x→0) e^x/2
=1/2
2)
lim(x→0) [1/ln(1+x)-1/x]
=lim(x→0) [x-ln(1+x)] /[xln(1+x)]
=lim(x→0) [x-ln(1+x)] /(x^2) (0----0型可导应用洛必达法则)
=lim(x→0) [1-1/(1+x)] /(2x)
=lim(x→0) 1/[2(1+x)]
=1/2
3)
lim(x→0) [e^x+e^(-x)-2] /(xsinx) (0---0型可导应用洛必达法则)
=lim(x→0) [e^x-e^(-x)] / (sinx+xcosx)
=lim(x→0) [e^x+e^(-x) / (cosx+cosx-xsinx)
=(1+1)/(1+1-0)
=1
1)
lim(x→0) [1/x-1/(e^x-1)]
=lim(x→0) (e^x-1-x)/[x(e^x-1)]
=lim(x→0) (e^x-1-x) /x^2 (0----0型可导应用洛必达法则)
=lim(x→0) (e^x-1)/(2x)
=lim(x→0) e^x/2
=1/2
2)
lim(x→0) [1/ln(1+x)-1/x]
=lim(x→0) [x-ln(1+x)] /[xln(1+x)]
=lim(x→0) [x-ln(1+x)] /(x^2) (0----0型可导应用洛必达法则)
=lim(x→0) [1-1/(1+x)] /(2x)
=lim(x→0) 1/[2(1+x)]
=1/2
3)
lim(x→0) [e^x+e^(-x)-2] /(xsinx) (0---0型可导应用洛必达法则)
=lim(x→0) [e^x-e^(-x)] / (sinx+xcosx)
=lim(x→0) [e^x+e^(-x) / (cosx+cosx-xsinx)
=(1+1)/(1+1-0)
=1
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