Question

设f(x)={1-cosx/x^2,x ＞ 0；a,x=0；bsinx/x,x＜0}试确定a与b的值， 使函数f(x)在x=0处连续。

Answer 1

当x＞0时，f(x)=(1-cosx)/x²=[1-1+2sin²(x/2)]/x²=[2sin²(x/2)]/x²
此时f(x)的左极限
=《x→0》lim f(x)
=《x→0》lim [2sin²(x/2)]/x²
=《x/2→0》lim [(1/2)sin²(x/2)]/(x/2)²
=《x/2→0》(1/2)lim [sin²(x/2)]/(x/2)²
=1/2

当x<0时，f(x)= bsinx/x
此时f(x)的右极限
=《x→0》lim f(x)
=《x→0》lim bsinx/x
=《x→0》blimsinx/x
=b

当x=0时f(x)=a

要使函数f(x)在x=0处连续，则只有左极限=右极限=f(0)，所以
a=b=1/2