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]内一致连续我知道要把问题归结到证明lim(x趋向于0+)f(x)存在,如何由lim(x趋向于0+)(√x)f`(x)存在导出lim(x趋向于0+)f(x)存在,
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/06 12:31:33
![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]内一致连续我知道要把问题归结到证明lim(x趋向于0+)f(x)存在,如何由lim(x趋向于0+)(√x)f`(x)存在导出lim(x趋向于0+)f(x)存在,](/uploads/image/z/3935927-47-7.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%E5%9C%A8%280%2C1%5D%E5%86%85%E8%BF%9E%E7%BB%AD%E5%8F%AF%E5%AF%BC%2C%E4%B8%94lim%28x%E8%B6%8B%E5%90%91%E4%BA%8E0%2B%29%28%E2%88%9Ax%29f%60%28x%29%E5%AD%98%E5%9C%A8%2C%E8%AF%81%E6%98%8Ef%28x%29%E5%9C%A8%280%2C1%5D%E5%86%85%E4%B8%80%E8%87%B4%E8%BF%9E%E7%BB%AD%E6%88%91%E7%9F%A5%E9%81%93%E8%A6%81%E6%8A%8A%E9%97%AE%E9%A2%98%E5%BD%92%E7%BB%93%E5%88%B0%E8%AF%81%E6%98%8Elim%28x%E8%B6%8B%E5%90%91%E4%BA%8E0%2B%29f%28x%29%E5%AD%98%E5%9C%A8%2C%E5%A6%82%E4%BD%95%E7%94%B1lim%28x%E8%B6%8B%E5%90%91%E4%BA%8E0%2B%29%28%E2%88%9Ax%29f%60%28x%29%E5%AD%98%E5%9C%A8%E5%AF%BC%E5%87%BAlim%28x%E8%B6%8B%E5%90%91%E4%BA%8E0%2B%29f%28x%29%E5%AD%98%E5%9C%A8%2C)
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]内一致连续我知道要把问题归结到证明lim(x趋向于0+)f(x)存在,如何由lim(x趋向于0+)(√x)f`(x)存在导出lim(x趋向于0+)f(x)存在,
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
我知道要把问题归结到证明lim(x趋向于0+)f(x)存在,如何由lim(x趋向于0+)(√x)f`(x)存在导出lim(x趋向于0+)f(x)存在,高手指点
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]内一致连续我知道要把问题归结到证明lim(x趋向于0+)f(x)存在,如何由lim(x趋向于0+)(√x)f`(x)存在导出lim(x趋向于0+)f(x)存在,
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.