1, The function f(x)=x+2 is continuous on R as a polynomial.
Therefore
x→climf(x)=f(c),c∈R Then
x→−1limf(x)=f(−1)=−1+2=1
2, The function f(x)=x−2 is continuous on R as a polynomial.
Therefore
x→climf(x)=f(c),c∈R Then
x→0limf(x)=f(0)=0−2=−2
3.
x→−1−limf(x)=(−1)2−1=0
x→−1+limf(x)=(−1−1)2−4=0 Since
x→−1−limf(x)=0=x→−1+limf(x), then x→−1limf(x) exists, and x→−1limf(x)=0.
4.
x→1−limf(x)=(1)3−1=0
x→1+limf(x)=(1)2+4=5 Since
x→1−limf(x)=0=5=x→1+limf(x), then x→1limf(x) does not exist.
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