Question

设奇函数f(x)在 [-a,a]上有定义a>0,在x=0处可导,问x=0是否为(x-sinx)f(x)/x^2的可去间断

Answer 1

lim(x-->0) (x-sinx)f(x)/x^2

= lim(x-->0) (x-sinx)/x^2 * lim(x-->0) f(x)

= f(0) * lim(x-->0) (1-cosx)/(2x) //洛必达法则

= f(0) * lim(x-->0) sinx /2 //洛必达法则

=0

极限存在.所以x=0是(x-sinx)f(x)/x^2的可去间断点.