证明f(x)=e的x次方-1(x>= 0)f(x)=0(x

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证明f(x)=e的x次方-1(x>= 0)f(x)=0(x
证明f(x)=e的x次方-1(x>= 0)f(x)=0(x<0)在x=0处极限的存在性

证明f(x)=e的x次方-1(x>= 0)f(x)=0(x
考虑左右极限,limx->0+ f(x) = limx->0+ e^x -1 = 0,limx->0- f(x) = limx->0- 0 = 0,所以极限为0.