因为：
| lim |
| x→0 |
| sin3x |
| x3 |
| f(x) |
| x2 |
| lim |
| x→0 |
| sin3x+xf(x) |
| x3 |
| lim |
| x→0 |
| ||
| x2 |
所以：
| lim |
| x→0 |
| sin3x |
| x |
又：f（x）在x=0的某领域内二阶可导，
所以：f（x），f′（x）在x=0连续，
从而：f（0）=-3．
由
| lim |
| x→0 |
| ||
| x2 |
得：
| lim |
| x→0 |
| ||
| x2 |
又易知：
| lim |
| x→0 |
3−
| ||
| x2 |
| lim |
| x→0 |
| 3x−sin3x |
| x3 |
| lim |
| x→0 |
| 3−3cos3x |
| 3x2 |
| lim |
| x→0 |
| 3sin3x |
| 2x |
| 9 |
| 2 |
故：
| lim |
| x→0 |
| f(x)+3 |
| x2 |
| 9 |
| 2 |
从而：f′(0)＝
| lim |
| x→0 |
| f(x)−f(0) |
| x−0 |
| lim |
| x→0 |
| f(x)+3 |
| x |
| lim |
| x→0 |
| f(x)+3 |
| x2 |
| 9 |
| 2 |
将f（x）在x=0处泰勒展开，并由
| lim |
| x→0 |
| f(x)+3 |
| x2 |
| 9 |
| 2 |
| lim |
| x→0 |
f(0)+f′(0)x+
| ||
| x2 |
| 9 |
| 2 |
计算得：
| 1 |
| 2 |