【答案解析】[解析] 记u
1
=sin u
0
,u
k+1
=sin u
k
,k=1,2,….
对
∈(-∞,+∞),k=1,2,…,有
f(u
0
)=f(sin u
0
)=f(u
1
)=f(sin u
1
)=f(sin u
2
)=…=f(sin u
k
)=f(u
k+1
),
即对
∈(-∞,+∞),n=1,2,…,都有f(u
0
)=f(u
n
)成立.
由于数列u
k
,k=1,2,…单调递减且有极限
=0.又f(x)在点x=0处连续,所以对
∈(-∞,+∞),f(u
0
)=