关于函数导数问题确定a、b的值,使函数f(x)=(1-cosax)/x,x0在（-∞ +∞）内处处可导f(x)是分段函数
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首先,要使f(x)在（-∞ +∞）内处处可导,须f(x)在（-∞ +∞）内连续
即须f(x)在x=0处连续
即lim(x->0-)f(x)=lim(x->0+)f(x)=f(0)
lim(x->0-)f(x)=lim(x->0-)[(1-cosax)/x]=lim(x->0-)asinax=0无论a为何值恒成立
lim(x->0+)f(x)=lim(x->0+)ln(b+x^2)/x=0,则
lim(x->0+)ln(b+x^2)=0,且lim(x->0+)[2x/(b+x^2)]=0
即b=1
其次要使f(x)在（-∞ +∞）内处处可导,只需要使f(x)在x=0处可导
即f'(0-)=f'(0+)
∵f'(0-)=d[(1-cosax)/x]/dx |(x=0)=lim(x->0)[axsinax-(1-cosax)]/x^2
=lim(x->0)[asinax+(a^2)xcosax-asinax]/2x
=lim(x->0)[(a^2)cosax-(a^3)xsinax]/2
=(a^2)/2
f'(0+)=d[ln(1+x^2)/x]/dx|(x=0)=lim(x->0)[2(x^2)/(1+x^2)-ln(1+x^2)]/x^2
=lim(x->0)[2(x^2)-(1+x^2)ln(1+x^2)]/[(x^2)(1+x^2)]
=lim(x->0)[4x-2xln(1+x^2)-2x]/(2x+4x^3)
=lim(x->0)[1-ln(1+x^2)]/(1+2x^2)
=1
∴(a^2)/2=1
a=±√2
综上可得
a=±√2, b=1
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