设f(x)在(x
0
-δ，x
0
+δ)有n阶连续导数，且f
(k)
(
0
)=0，k=2，3，…，n-1；f
(n)
(x
0
)≠0．当0＜｜h｜＜δ时，f(x
0
+h)=f(x
0
)=hf'(x
0
+θh)，(0＜θ＜1)．求证：
【正确答案】正确答案：这里m=1，求的是f(x
0
+h)-f(x
0
)=hf'(x
0
+θh)(0＜θ＜1)当h→0时中值θ的极限．为解出θ，按题中条件，将f'(x
0
+θh)在x=x
0
展开成带皮亚诺余项的n-1阶泰勒公式得 f'(x
0
+θh)=f'(x
0
)+f''(x
0
)θh+
f
(3)
(x
0
)(θh)
2
+…+
f
(n)
(x
0
)(θh)
n-1
+o(h
n-1
) =f'(x
0
)+
f
(n)
(x
0
)(θh)
n-1
+o(h
n-1
)(h→0)， 代入原式得 (x
0
+h)-f(x
0
)=hf'(x
0
)+
f
(n)
(x
0
)θ
n-1
h
n
+o(h
n
) ① 再将f(x
0
+h)在x=x
0
展开成带皮亚诺余项的n阶泰勒公式 f(x
0
+h)-f(x
0
)=f'(x
0
)h+…+
f
(n)
(x
0
)h
b
+o(h
n
) =f'(x
0
)h+
f
(n)
(x
0
)h
n
+o(h
n
)(h→0)， ② 将②代入①后两边除以h
n
得
令h→0，得
f
(3)
(x
0
)(θh)
2
+…+
f
(n)
(x
0
)(θh)
n-1
+o(h
n-1
) =f'(x
0
)+
f
(n)
(x
0
)(θh)
n-1
+o(h
n-1
)(h→0)， 代入原式得 (x
0
+h)-f(x
0
)=hf'(x
0
)+
f
(n)
(x
0
)θ
n-1
h
n
+o(h
n
) ① 再将f(x
0
+h)在x=x
0
展开成带皮亚诺余项的n阶泰勒公式 f(x
0
+h)-f(x
0
)=f'(x
0
)h+…+
f
(n)
(x
0
)h
b
+o(h
n
) =f'(x
0
)h+
f
(n)
(x
0
)h
n
+o(h
n
)(h→0)， ② 将②代入①后两边除以h
n
得
令h→0，得
【答案解析】