设f''(x)连续,且f''(x)>0,f(0)=f'(0)=0,试求极限lim(x->0+)∫(上u(x) 下0)f(t)dt/∫(上x下0)f(t)dt其中u(x)是曲线y=f(x)在点(x,f(x))处的切线在x轴上的截距
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![设f''(x)连续,且f''(x)>0,f(0)=f'(0)=0,试求极限lim(x->0+)∫(上u(x) 下0)f(t)dt/∫(上x下0)f(t)dt其中u(x)是曲线y=f(x)在点(x,f(x))处的切线在x轴上的截距](/uploads/image/z/11630466-18-6.jpg?t=%E8%AE%BEf%27%27%28x%29%E8%BF%9E%E7%BB%AD%2C%E4%B8%94f%27%27%28x%29%3E0%2Cf%280%29%3Df%27%280%29%3D0%2C%E8%AF%95%E6%B1%82%E6%9E%81%E9%99%90lim%EF%BC%88x-%3E0%2B%29%E2%88%AB%28%E4%B8%8Au%28x%29+%E4%B8%8B0%EF%BC%89f%28t%29dt%2F%E2%88%AB%28%E4%B8%8Ax%E4%B8%8B0%EF%BC%89f%28t%29dt%E5%85%B6%E4%B8%ADu%28x%29%E6%98%AF%E6%9B%B2%E7%BA%BFy%3Df%28x%29%E5%9C%A8%E7%82%B9%28x%2Cf%28x%29%29%E5%A4%84%E7%9A%84%E5%88%87%E7%BA%BF%E5%9C%A8x%E8%BD%B4%E4%B8%8A%E7%9A%84%E6%88%AA%E8%B7%9D)
设f''(x)连续,且f''(x)>0,f(0)=f'(0)=0,试求极限lim(x->0+)∫(上u(x) 下0)f(t)dt/∫(上x下0)f(t)dt其中u(x)是曲线y=f(x)在点(x,f(x))处的切线在x轴上的截距
设f''(x)连续,且f''(x)>0,f(0)=f'(0)=0,试求极限lim(x->0+)∫(上u(x) 下0)f(t)dt/∫(上x下0)f(t)dt
其中u(x)是曲线y=f(x)在点(x,f(x))处的切线在x轴上的截距
设f''(x)连续,且f''(x)>0,f(0)=f'(0)=0,试求极限lim(x->0+)∫(上u(x) 下0)f(t)dt/∫(上x下0)f(t)dt其中u(x)是曲线y=f(x)在点(x,f(x))处的切线在x轴上的截距
先求出u(x) = f(x) - xf'(x)
u' = -xf''(x)
对原式用洛必达法则得
=-f(u)*x*f''(x) / f(x)
由于f''(x) > 0
求xf(u)/f(x)的极限,使用洛必达法则得
=[f(u)+xf'(u)*u'] / f'(x)
再用洛必达法则
=[f'(u)*u' + f'(u)*u' + x * u' * f''(u) * u' + xf'(u) * u''] / f''(x)
=0
所以原式 = 0
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