设函数f（x）在x=0点的某个邻域内连续，且limx→0f(x)ex?1=2，则曲线y=f（x）在x=0处的法线方程为 设函数f（x），g（x）在x=0的某个邻域内连续，且limx...
\u5df2\u77e5f(x)\u5728x=0\u7684\u67d0\u4e2a\u90bb\u57df\u5185\u8fde\u7eed,\u4e14limx->0f(x)/1-cosx=2,\u5219\u5728x=0\u5904\u8bc1\u660e\uff1a\u7531(x\u21920)limg(x)/x=-1 \uff08\u6781\u9650\u4e3a-1,\u5206\u6bcd\u8d8b\u4e8e0,\u5219\u5206\u5b50\u5fc5\u8d8b\u4e8e0\uff09
\u53ef\u77e5(x\u21920)limg(x)=0 \u5373g(0)=0
\u4e8e\u662f(x\u21920)lim[g(x)-g(0)]/(x-0)=-1
\u5219g(x)\u5728\u8be5\u90bb\u57df\u5185\u53ef\u5bfc\u4e14g'(0)=-1
(x\u21920)limf(x)/g²(x)=2
\u56e0\u4e3a(x\u21920)limg²(x)=0
\u5219(x\u21920)limf(x)=0
f(0)=0
\u5bf9(x\u21920)limf(x)/g²(x)=2\u8fdb\u884c\u53d8\u5f62
(x\u21920)limf(x)/g²(x)
=(x\u21920)lim[f(x)/x][x²/g(x)]
=(x\u21920)lim[f(x)/x²]•(x\u21920)limx²/g(x) \uff08\u53d8\u6210\u4e24\u4e2a\u6781\u9650\u4e4b\u79ef,\u5e76\u5bf9\u53f3\u8fb9\u7684\u6781\u9650\u7528\u6d1b\u5fc5\u8fbe\u6cd5\u5219\uff09
=(x\u21920)lim[f(x)/x²]•(x\u21920)limx/g(x)•(x\u21920)lim1/g'(x)
=(x\u21920)lim[f(x)/x²]•(-1)•(-1)
=2
\u56e0\u6b64f(x)=2x²+o(x)
\u4e8e\u662f\u53ef\u4ee5\u5f97\u5230(x\u21920)limf(x)/x=0
\u5373f'(0)=0
\u7531\u4e8e1-cosx\uff5e12x2\uff08x\u21920\uff09\uff0c\u56e0\u6b64\uff0c\u7531limx\u21920g(x)1?cosx\uff1d1\u5f97\u5230g\uff08x\uff09\uff5e12x2\uff08x\u21920\uff09\uff0c\u2234\u7531limx\u21920f(x)g2(x)=2\uff0c\u5f97limx\u21920f(x)2g2(x)\uff1dlimx\u21920f(x)12x4\uff1d1\uff0c\u4e14limx\u21920f(x)\uff1d0=f\uff080\uff09\u53c812x4\u22650\uff0c\u56e0\u800c\u5728x=0\u7684\u67d0\u4e2a\u90bb\u57df\u5185f\uff08x\uff09\u22650\u2234x=0\u662ff\uff08x\uff09\u7684\u6781\u5c0f\u503c\u70b9\u6781\u503c\u5c31\u662ff\uff080\uff09=0
因为：
| lim |
| x→0 |
| f(x) |
| ex?1 |
| lim |
| x→0 |
所以：f（0）=
| lim |
| x→0 |
利用导数的定义可得：
f′（0）=
| lim |
| x→0 |
| f(x)?f(0) |
| x?0 |
| lim |
| x→0 |
| f(x) |
| x |
| lim |
| x→0 |
| f(x) |
| ex?1 |
| ex?1 |
| x |
| lim |
| x→0 |
| f(x) |
| ex?1 |
| lim |
| x→0 |
| ex?1 |
| x |
所以，y=f（x）在x=0的切线的斜率为2，
故：法线斜率为?
| 1 |
| 2 |
从而，曲线y=f（x）在x=0处的法线方程为：
y-f（0）=?
| 1 |
| 2 |
即：y=?
| 1 |
| 2 |
故答案为：y=?
| 1 |
| 2 |
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