设f(x)=g(x)/x(x不等于0),f(x)=0(x=0),且已知g(0)=g'(0)=0,g''(0)=3,试求f ' (0)?
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limx->0 f(x) =limx->0 g(x)/x (∵g(0)=0 则是0/0型,用洛必达)
=limx->0 g'(x)/1
=g'(0)/1
=0
=f(0)
∴f(x)在点x=0连续
∴
f ' (x)=[g(x)/x]'
=[g'(x)x-g(x)x']/x^2
=[g'(x)x-g(x)]/x^2
f'(0)=[g'(0)*0-g(0)]/0^2 (这个不通,只有用极限了)
limx->0 f'(x)=limx->0 [g'(x)x-g(x)]/x^2 (0/0型,用洛必达)
=limx->0 [g''(x)x+g'(x) -g'(x)] /2x
=limx->0 g''(x) /2
=g''(0)/2
=3/2
∴f'(0)=3/2
=limx->0 g'(x)/1
=g'(0)/1
=0
=f(0)
∴f(x)在点x=0连续
∴
f ' (x)=[g(x)/x]'
=[g'(x)x-g(x)x']/x^2
=[g'(x)x-g(x)]/x^2
f'(0)=[g'(0)*0-g(0)]/0^2 (这个不通,只有用极限了)
limx->0 f'(x)=limx->0 [g'(x)x-g(x)]/x^2 (0/0型,用洛必达)
=limx->0 [g''(x)x+g'(x) -g'(x)] /2x
=limx->0 g''(x) /2
=g''(0)/2
=3/2
∴f'(0)=3/2
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