设函数在x=0处连续且x趋于0时lim f(2x)/3x=1求过点(0,f(0))的切线方程
人气:478 ℃ 时间:2019-09-29 00:13:41
解答
由已知lim f(2x)/3x=1,得 f(0)=0 (否则极限不存在)
且有 lim f(2x)/3x=1=lim [f(2x)-f(0)]/[2x-0] *(2/3)=2/3*f'(0)
=>f'(0)=3/2
所以函数在(0,f(0))即(0,0)点的导数为3/2,则切线方程为y=3/2 * x.
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