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Identity if the limit of f(x)f\left(x\right)f(x) exists given the LHL (left-hand limit) and RHL (right-hand limit).
limx→c+f(x)=0\lim_{x\to c+f\left(x\right)=0}limx→c+f(x)=0 Limit Exist
limx→c−f(x)=0\lim_{x\to c-f\left(x\right)=0}limx→c−f(x)=0 Limit Exist
limx→c+f(x)=5\lim_{x\to c+f\left(x\right)=5}limx→c+f(x)=5 Limit Exist
limx→c−f(x)=5\lim_{x\to c-f\left(x\right)=5}limx→c−f(x)=5 Limit does not exist
limx→c+f(x)=∞\lim_{x\to c+f\left(x\right)=\infty}limx→c+f(x)=∞ Limit Exist
limx→c−f(x)=∞\lim_{x\to c-f\left(x\right)=\infty}limx→c−f(x)=∞ Limit does not exist
limx→c+f(x)=2\lim_{x\to c+f\left(x\right)=2}limx→c+f(x)=2 Limit Exist
limx→c−f(x)=2\lim_{x\to c-f\left(x\right)=2}limx→c−f(x)=2 Limit does not exist
limx→c+f(x)=−4\lim_{x\to c+f\left(x\right)=-4}limx→c+f(x)=−4 Limit Exist
limx→c+f(x)=1\lim_{x\to c+f\left(x\right)=1}limx→c+f(x)=1 Limit Exist
limx→c−f(x)=1\lim_{x\to c-f\left(x\right)=1}limx→c−f(x)=1 Limit does not exist
limx→c+f(x)=3\lim_{x\to c+f\left(x\right)=3}limx→c+f(x)=3 Limit Exist
limx→c−(f)=4\lim_{x\to c-\left(f\right)=4}limx→c−(f)=4 Limit does not exist
limx→c+f(x)=−∞\lim_{x\to c+f\left(x\right)=-\infty}limx→c+f(x)=−∞ Limit Exist
limx→c−f(x)=4\lim_{x\to c-f\left(x\right)=4}limx→c−f(x)=4 Limit does not exist
It is done.