设函数f(x)连续,且limx→0f(x)−sinxx
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设函数f(x)连续,且
lim |
x→0 |
f(x)−sinx |
x |
因为
lim
x→0
f(x)−sinx
x=a(a为常数),
所以
lim
x→0f(x)−sinx=0,
故 f(0)=
lim
x→0f(x)=0.
当x=0时,
F(x)=
∫10f(xy)dy
=
∫10f(0)dy
=0;
∀x≠0,令t=xy,则dt=xdy,
从而,
dy=
dt
x,
F(x)=
∫10f(xy)dy
=
∫x0
f(t)
xdt
=
1
x
∫ x0f(t)dt,
故 F(x)=
1
x
∫ x0f(t)dt, x≠0
0, x=0.
因为f(x)连续,所以
lim
x→0F(x)=
lim
x→0
1
x
∫ x0f(t)dt
=
lim
x→0
(
∫x0f(t)dt)′
x′
=
lim
x→0f(x)
=f(0)=0.
从而,F(0)=
lim
x→0
lim
x→0
f(x)−sinx
x=a(a为常数),
所以
lim
x→0f(x)−sinx=0,
故 f(0)=
lim
x→0f(x)=0.
当x=0时,
F(x)=
∫10f(xy)dy
=
∫10f(0)dy
=0;
∀x≠0,令t=xy,则dt=xdy,
从而,
dy=
dt
x,
F(x)=
∫10f(xy)dy
=
∫x0
f(t)
xdt
=
1
x
∫ x0f(t)dt,
故 F(x)=
1
x
∫ x0f(t)dt, x≠0
0, x=0.
因为f(x)连续,所以
lim
x→0F(x)=
lim
x→0
1
x
∫ x0f(t)dt
=
lim
x→0
(
∫x0f(t)dt)′
x′
=
lim
x→0f(x)
=f(0)=0.
从而,F(0)=
lim
x→0
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