用函数极限的“ε-δ”定义证明:lim┬(x→π/4)sinx =√2/2
用函数极限的“ε-δ”定义证明:lim┬(x→π/4)sinx =√2/2
数学人气:709 ℃时间:2020-02-05 18:41:08
优质解答
对于任意的ε>0,取δ=ε,则当|x-π|
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