应该是证g(x)在R上有一阶连续导数吧?
当x≠0时,g(x)=f(x)/x
∴g'(x) = [xf'(x)-f(x)]/x²
g'(x)在x≠0时连续
x=0时,
g'(0) = lim(x→0) [g(x)-g(0)]/(x-0)
=lim(x→0) [f(x)/x-f'(0)]/x
=lim(x→0) [f(x)-xf'(0)]/x²
=lim(x→0) [f'(x)-f'(0)]/(2x)
=(1/2)f''(0)
又lim(x→0) [xf'(x)-f(x)]/x²
=lim(x→0) [f'(x)+xf''(x)-f'(x)]/(2x)
=(1/2)f''(0)
∴lim(x→0) g'(x) =g'(0)
即g'(x)在x=0处连续
综上可得g'(x)在R上连续,即g(x)在R上有一阶连续导数