设函数f(x)在(-1,1)有定义且满足x≤f(x)≤x²+x证明f'(0)存在且f'(0)=1

Question

设函数f(x)在(-1,1)有定义且满足x≤f(x)≤x²+x证明f'(0)存在且f'(0)=1

Answer 1

证明：由已知,取x=0,得 0≤f(x)≤0
=>f(0)=0.
由当x->0+时,1=lim x/x≤ limf(x)/x ≤ lim (x^2+x)/x=1
=>limf(x)/x=1 (夹逼准则)
由当x->0-时,1=lim (x^2+x)/x ≤ limf(x)/x ≤ lim x/x=1
=>limf(x)/x=1 (夹逼准则)
所以有,x->0时,有 limf(x)/x=1=lim(f(x)-f(0))/(x-0)=f'(0)
即f'(0)存在且f'(0)=1.