EX 5.設f
(
x
)
=
{
g
(
x
)
sin 1
x, x ≠ 0
0, x=0
, g
(
0
)
=g
'
(
0
)
=0,試求 f
'
(
0
)
g
'
(
0
)
=lim
x →0
g
(
x
)
−g
(
0
)
x=lim
x →0
g
(
x
)
x=0
⇒f'
(
0
)
=lim
x→ 0
f
(
x
)
−f
(
0
)
x=lim
x →0
g
(
x
)
sin 1
x
x
∵0≤
|
g
(
x
)
sin 1
x
x
|
≤
|
g
(
x
)
x
|
且lim
x → 0
0=lim
x→ 0
|
g
(
x
)
x
|
=0
∴由夾擠定理知lim
x→ 0
|
g
(
x
)
sin 1
x
x
|
=0⇒lim
x → 0
g
(
x
)
sin 1
x
x=f'
(
0
)
=0
