设函数f(x)=x^2+1 x<0,f(x)=2X+1 x>=0 则极限lim(x→0)f(x)=?
人气:300 ℃ 时间:2019-08-21 01:02:01
解答
f(x) = { x² + 1,x < 0
{ 2x + 1,x ≥ 0
lim(x→0⁻) f(x) = lim(x→0⁻) (x² + 1) = 0 + 1 = 1
lim(x→0⁺) f(x) = lim(x→0⁺) (2x + 1) = 2(0) + 1 = 1
∴lim(x→0) f(x) = 1
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