Calculus 10th Edition

For this piece-wise function, the only possible point of discontinuity is $x=1\to$ $\lim\limits_{x\to1^+}f(x)=(1^+)^2=1;\lim\limits_{x\to1^-}f(x)=(1^-)=1.$ Since $\lim\limits_{x\to1^+}f(x)=;\lim\limits_{x\to1^-}f(x)\to;\lim\limits_{x\to1}f(x)=1.$ Furthermore the limit is equal to the value of the function at that point hence the function is continuous at $x=1$ and does not have any discontinuities at all.