Since you did not provide a picture with a graph, then I can show the possible options.
In this case,
x→−2−limf(x)=1x→−2+limf(x)=4x→−2−limf(x)=x→−2+limf(x)⟶∄x→−2limf(x)f(x) is discontinuous at the pointx=−2x→2−limf(x)=4x→2+limf(x)=4x→2−limf(x)=x→2+limf(x)⟶x→2limf(x)=4f(x) is continuous at the pointx=2
Another situation
x→0−limf(x)=1x→0+limf(x)=1x→0−limf(x)=x→0+limf(x)⟶x→0limf(x)=1f(0)=0=x→0limf(x)=1f(x) is discontinuous at the pointx=0 Another situation
x→−2−limf(x)=0x→−2+limf(x)=+∞x→−2−limf(x)=x→−2+limf(x)⟶∄x→−2limf(x)f(x) is discontinuous at the pointx=−2
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