例1证明函数y=2x+1在x=1处连续 证Δy=2(x+△x)+1-[2x+1 2△v ∴当Δx→0时,Δy→>0 ∴函数p=2x+1在x=1处连续 Economic--mathematics 19-6 Wednesday, February 24, 2021Economic--mathematics 19 - 6 Wednesday, February 24, 2021 例1 证明函数 y = 2x＋1在x = 1处连续. 证 y = 2(x + x)＋1− 2x＋1 = 2x 当x → 0时,y → 0.  函数 y = 2x＋1在x = 1处连续