设函数f(x)=x^2+1 x=0 则极限lim(x→0）f（x)=?

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