Question

设f（x）在x=0点的某邻域内连续，且f（0）=1，则limx→0∫x0dt∫ t0f(u)dusinx2=1212

设f（x）在x=0点的某邻域内连续，且f（0）=1，则limx→0∫x0dt∫ t0f(u)dusinx2=1212．

Answer 1

lim

x→0

∫ x 0 dt ∫  t 0 f(u)du

sinx2

=

lim

x→0

( ∫  x 0 dt ∫  t 0 f(u)du)′

(sinx2)′

=

lim

x→0

∫ x 0 f(u)du

2xcosx2

=

lim

x→0

∫ x 0 f(u)du

2x

=

lim

x→0

( ∫  x 0 f(u)du)′

(2x)′

=

lim

x→0

f(x)

2

=

f(0)

2

=

1

2

．
故答案为：

1

2

．