设函数f(x)=x^2+1 x=0 则极限lim(x→0)f(x)=?
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![设函数f(x)=x^2+1 x=0 则极限lim(x→0)f(x)=?](/uploads/image/z/5222542-22-2.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3Dx%5E2%2B1+x%3D0+%E5%88%99%E6%9E%81%E9%99%90lim%28x%E2%86%920%EF%BC%89f%EF%BC%88x%29%3D%3F)
设函数f(x)=x^2+1 x=0 则极限lim(x→0)f(x)=?
设函数f(x)=x^2+1 x<0,f(x)=2X+1 x>=0 则极限lim(x→0)f(x)=?
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
左右极限都为1,故极限为1