设f(x)在点x0处可导，计算极限 设函数f（x）在点x0处可导，试求下列各极限的值．（1）li...

\u8bbe\u51fd\u6570f\uff08x\uff09\u5728\u70b9x 0 \u5904\u53ef\u5bfc\uff0c\u8bd5\u6c42\u4e0b\u5217\u5404\u6781\u9650\u7684\u503c\u3002\uff081\uff09 \uff1b\uff082\uff09 \u3002

\u89e3\uff1a\uff081\uff09\u539f\u5f0f= =-f\u2032\uff08x 0 \uff09\uff08\u25b3x\u21920\u65f6\uff0c-\u25b3x\u21920\uff09 \uff082\uff09\u539f\u5f0f= \u3002

\uff081\uff09\u539f\u5f0f=lim\u25b3x\u21920f(x0?\u25b3x)?f(x0)?(?\u25b3x)=-lim\u25b3x\u21920f(x0?\u25b3x)?f(x0)?\u25b3x=-f\u2032\uff08x0\uff09\uff082\uff09limh\u21920f(x0+h)?f(x0?h)2h=12limh\u21920 f(x0+h)?f(x0)+f(x0)?f(x0?h)h=12limh\u21920[f(x0+h)?f(x0)h?f(x0?h)?f(x0)?h]=12[f\u2032\uff08x0\uff09+f\u2032\uff08x0\uff09]=f\u2032\uff08x0\uff09\uff0e

原式=lim[xf(x0)-xf(x)+xf(x)-x0f(x)]/(x-x0)
=lim[xf(x0)-xf(x)]/(x-x0)+lim[xf(x)-x0f(x)]/(x-x0)
=limx[f(x0)-f(x)]/(x-x0)+f(x0)
=-x0f'(x0)+f(x0)
=f(x0)-x0f'(x0)

洛必达法则
由题意知，当X趋近X0时，分子和分母都趋近与0
根据洛必达法则 此时函数极限=分子导数/分母导数
xf(x0)-x0f(x)导数=f(x0)-x0f'(x)
x-x0的导数=1
所以结果就是f(x0)-x0f'(x)

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