f(x)具有二阶连续导数,且
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f(x)具有二阶连续导数,且
f(0)=f'(0)=0,f''(x)>0,u(x)是y=f(x)在点(x,f(x))处的切线在x轴上的截距,则limx趋于0x/u(x)=?
f(0)=f'(0)=0,f''(x)>0,u(x)是y=f(x)在点(x,f(x))处的切线在x轴上的截距,则limx趋于0x/u(x)=?
![f(x)具有二阶连续导数,且](/uploads/image/z/19166480-8-0.jpg?t=f%28x%29%E5%85%B7%E6%9C%89%E4%BA%8C%E9%98%B6%E8%BF%9E%E7%BB%AD%E5%AF%BC%E6%95%B0%2C%E4%B8%94)
x0处切线为y=f(x0)+f'(x0)(x-x0)
所以u(x0)=x0-f(x0)/f'(x0)
即u(x)=x-f(x)/f'(x)
所以lim(x→0)x/u(x)
=lim(x→0)xf'(x)/(xf'(x)-f(x))
=lim(x→0)(f'(x)+xf''(x))/(f'(x)+xf''(x)-f'(x)) (洛必达法则)
=lim(x→0)(f'(x)+xf''(x))/(xf''(x))
=lim(x→0)(f'(x)/x+f''(x))/f''(x)
=lim(x→0)[(f'(x)-f'(0))/(x-0)+f''(x)]/f''(x)
=(f''(0)+f''(0))/f''(0)
=2
所以u(x0)=x0-f(x0)/f'(x0)
即u(x)=x-f(x)/f'(x)
所以lim(x→0)x/u(x)
=lim(x→0)xf'(x)/(xf'(x)-f(x))
=lim(x→0)(f'(x)+xf''(x))/(f'(x)+xf''(x)-f'(x)) (洛必达法则)
=lim(x→0)(f'(x)+xf''(x))/(xf''(x))
=lim(x→0)(f'(x)/x+f''(x))/f''(x)
=lim(x→0)[(f'(x)-f'(0))/(x-0)+f''(x)]/f''(x)
=(f''(0)+f''(0))/f''(0)
=2
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