∫f(x)dx=3e^(x/3)-x+c,则 lim f(x)/x= (x趋于0)

来源:学生作业帮助网 编辑:作业帮 时间:2024/04/27 22:14:15
∫f(x)dx=3e^(x/3)-x+c,则 lim f(x)/x= (x趋于0)

∫f(x)dx=3e^(x/3)-x+c,则 lim f(x)/x= (x趋于0)
∫f(x)dx=3e^(x/3)-x+c,则 lim f(x)/x= (x趋于0)

∫f(x)dx=3e^(x/3)-x+c,则 lim f(x)/x= (x趋于0)
∫f(x)dx=3e^(x/3)-x+c
两边求导得
f(x)=e^(x/3)-1
f(0)=0
lim(x→0) f(x)/x
=lim(x→0) [f(x)-f(0)]/(x-0)
=f'(0)
=1/3e^(x/3)
=1/3

∫f(x)dx=3e^(x/3)-x+c
求导
f(x)=e^(x/3)-1
x趋于0
则e^(x/3)-1~x/3
所以lim f(x)/x=lim (x/3)/x=1/3

∫f(x)dx=3e^(x/3)-x+c
f(x) = e^(x/3)-1
lim(x->0) f(x)/x
=lim(x->0) (e^(x/3)-1)/x (0/0)
=lim(x->0)(1/3)e^(x/3)/1
=1/3

∫f(x)dx=3e^(x/3)-x+c
f(x)=[3e^(x/3)-x+c]'=e^(x/3)-1
lim f(x)/x= lim [e^(x/3)-1]/x=lim[(1/3)e^(x/3)]/1(洛必达法则)=1/3

f(x)=(3e^(x/3)-x+c)'=e^(x/3)-1
lim f(x)/x=lim[e^(x/3)-1]/x
=lime^(x/3)/3 罗必塔法则
=1/3

f(x)=[3e^(x/3)-x+c]'=e^(x/3)-1
f(x)/x x趋于0,分子,分母都趋于零,可用洛必达法则,上下求一次导
[f(x)]'=(1/3)e^(x/3)
[x]'=1
极限为1/3