设函数f(x)在(0,1]内连续可导,且lim(x趋向于0+)(√x)f`(x)存在,证明f(x)在(0,1]内一致连续

个人认为没必要先证limf(x)存在,将其作为一致连续性的推论更合适(用Cauchy收敛准则).
f'(x)在(0,1]连续,lim(√x)f'(x)存在,可得(√x)f'(x)在(0,1]有界,设有|(√x)f'(x)| < M.
对任意a,b∈(0,1],a < b,在[a,b]上对f(x)与√x使用Cauchy中值定理得:存在c∈(a,b)使
(f(a)-f(b))/(√a-√b) = f'(c)/(1/(2√c)) = 2(√c)f'(c).
于是|f(a)-f(b)| = 2(√c)f'(c)|√a-√b| < 2M|√a-√b|.
又√x在[0,1]连续故一致连续:对任意ε > 0,存在δ > 0使当|a-b| < δ时有|√a-√b| < ε/(2M).
则|a-b| < δ时,|f(a)-f(b)| < 2M|√a-√b| < ε.
即我们证明了f(x)一致连续.
其实微调一下证法,命题的条件可以减弱为f(x)在(0,1]连续,在某个(0,δ)内可导且(√x)f'(x)有界.