(f(arctanx)-f(π/4))/(x-1)在x趋向1时的极限为1,函数f(u)可微,求f'(π/4)
问题描述：
(f(arctanx)-f(π/4))/(x-1)在x趋向1时的极限为1,函数f(u)可微,求f'(π/4)
答
(f(arctanx)-f(π/4))/(x-1)在x趋向1时的极限为1
即
lim(x->1) (f(arctanx)-f(π/4))/(arctanx-π/4)· (arctanx-π/4)/(x-1)=1
f'(π/4)·lim(x->1)(arctanx-π/4)/(x-1)=1
f'(π/4)·lim(x->1)(1/(1+x²))/1=1
f'(π/4)·1/2=1
f'(π/4)=2