证明:当x>0时,1/x>arctanx-π/2

问题描述:

证明:当x>0时,1/x>arctanx-π/2

设f(x)=1/x-arctanx-π/2
f′=-1/x²-1/(1+x²)0
即1/x>arctanx-π/2为啥“ 当x趋于+∞时,f(x)趋于0 ; 所以,f(x)>0”?f′=-1/x²-1/(1+x²)a,f(b)