用洛必达法则计算极限:limx→0 (e^x-cosx)/(sinx)lim x→π (sin3x)/(tan5x);lim x→0 (e^x+e^[-x]-2)/(1-cosx) ;lim x→0 [x(e^x+1)-2(e^x-1)]/x^3 ;lim x→π/2 (Insinx)/(π-2x)^2

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用洛必达法则计算极限:limx→0(e^x-cosx)/(sinx)limx→π(sin3x)/(tan5x);limx→0(e^x+e^[-x]-2)/(1-cosx);limx→0[x(e^x+1

用洛必达法则计算极限:limx→0 (e^x-cosx)/(sinx)lim x→π (sin3x)/(tan5x);lim x→0 (e^x+e^[-x]-2)/(1-cosx) ;lim x→0 [x(e^x+1)-2(e^x-1)]/x^3 ;lim x→π/2 (Insinx)/(π-2x)^2
用洛必达法则计算极限:limx→0 (e^x-cosx)/(sinx)
lim x→π (sin3x)/(tan5x);lim x→0 (e^x+e^[-x]-2)/(1-cosx) ;lim x→0 [x(e^x+1)-2(e^x-1)]/x^3 ;lim x→π/2 (Insinx)/(π-2x)^2

用洛必达法则计算极限:limx→0 (e^x-cosx)/(sinx)lim x→π (sin3x)/(tan5x);lim x→0 (e^x+e^[-x]-2)/(1-cosx) ;lim x→0 [x(e^x+1)-2(e^x-1)]/x^3 ;lim x→π/2 (Insinx)/(π-2x)^2
0/0型,分数上下求导,得:e^x+sinx/cosx = 1