## Calculus 10th Edition

For this piece-wise function, the only possible point of discontinuity is $x=1:$ $\lim\limits_{x\to1^+}f(x)=(1^+)^2=1;\lim\limits_{x\to1^-}f(x)=-2(1^-)+3=1.$ 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 function's value at the point $x=1$ hence the function is continuous at $x=1$ and has no discontinuities at all.