设f（x）={2(1-cosx)/x^2,x0讨论f(x)在R内的可导性

Question

设f（x）={2(1-cosx)/x^2,x0 讨论f(x)在R内的可导性

Answer 1

x0时,f(x)为含参积分,任然处处可导.只需考虑x=0的情况.
f(x)在0处的左极限为
lim{x→0-} f(x)
= lim{x→0-} 2(1-cosx)/x^2
= lim{x→0-} 2(x^2/2)/x^2 = 1.
f(x)在0处的右极限为
lim{x→0+} f(x)
= lim{x→0+} 1/x∫(0,x)cost^2dt
根据罗必塔法则,有
lim{x→0+} 1/x∫(0,x)cost^2dt
= lim{x→0+} d(∫(0,x)cost^2dt)/dx
= lim{x→0+} cosx^2 = 1.
从而lim{x→0-} f(x) = f(0) = lim{x→0+} f(x) ,即,f(x)在x=0处连续.
接下来,f(x)在0处的左导数为
lim{x→0-} (f(x) - f(0))/x
= lim{x→0-} (2(1-cosx)/x^2 - 1)/x
= lim{x→0-} (2(1-cosx)-x^2)/x^3
= lim{x→0-} (2sinx-2x)/3x^2 (罗必塔法则)
= lim{x→0-} (2cosx-2)/6x (罗必塔法则)
= lim{x→0-} (-x^2)/6x (cos x -1 为 -x^2/2等价无穷小)
= 0.
而f(x)在0处的右导数为
lim{x→0+} (f(x) - f(0))/x
= lim{x→0+} (1/x∫(0,x)cost^2dt - 1)/x
= lim{x→0+} (∫(0,x)cost^2dt - x)/x^2 (罗必塔法则)
= lim{x→0+} (cosx^2 - 1)/2x (罗必塔法则)
= lim{x→0+} (-2x sinx^2 )/2 (罗必塔法则)
= 0.
可见f(x)在x=0处左右导数相等,同时因为f(x)在x=0处连续,所以f(x)在x=0处可导.
综上,f(x)在R上处处可导.