用洛必达法则求两道题的极限~1.lim(x→0)(e^x-1)/(sinx)2.lim(x→0)〔In(1+x)-x〕/(cosx-1)

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用洛必达法则求两道题的极限~1.lim(x→0)(e^x-1)/(sinx)2.lim(x→0)〔In(1+x)-x〕/(cosx-1)
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用洛必达法则求两道题的极限~1.lim(x→0)(e^x-1)/(sinx)2.lim(x→0)〔In(1+x)-x〕/(cosx-1)
用洛必达法则求两道题的极限~
1.lim(x→0)(e^x-1)/(sinx)
2.lim(x→0)〔In(1+x)-x〕/(cosx-1)

用洛必达法则求两道题的极限~1.lim(x→0)(e^x-1)/(sinx)2.lim(x→0)〔In(1+x)-x〕/(cosx-1)
1,分子分母同时求导
lim(x→0)(e^x-1)/(sinx)
=lim(x→0)e^x/cosx=1
2,lim(x→0)〔In(1+x)-x〕/(cosx-1)
=lim(x→0)〔1/(1+x)-1)/(-sinx)
=lim(x→0)[1/(1+x)]*x/sinx
=lim(x→0)x/sinx
=lim(x→0)1/cosx
=1

1 lim(x→0) (e^x-1)/(sinx)=lim(x→0) x/x=1
2 lim(x→0)〔In(1+x)-x〕/(cosx-1)
=lim(x→0)[1/(1+x)-1]/-sinx
=lim(x→0)(x/1+x)/sinx
=lim(x→0) (x/1+x)/x
=lim(x→0) 1/1+x
=1