## Calculus 10th Edition

For this piece-wise function, there are two points where a possible discontinuity could occur: $x=1$ or $x=-1$ $\lim\limits_{x\to1^-}f(x)=\tan{\dfrac{\pi(1^-)}{4}}=1;\lim\limits_{x\to1^+}f(x)=(1^+)=1.$ Since $\lim\limits_{x\to1^-}f(x)=\lim\limits_{x\to1^+}f(x)\to\lim\limits_{x\to1}f(x)=1=f(1)$ hence the function is continuous at $x=1.$ $\lim\limits_{x\to-1^-}f(x)=(-1^-)=-1; \lim\limits_{x\to-1^+}f(x)=\tan{\dfrac{\pi(-1^+)}{4}}=-1.$ Since $\lim\limits_{x\to-1^-}f(x)=\lim\limits_{x\to-1^+}f(x)\to\lim\limits_{x\to-1}f(x)=-1=f(-1)$ hence the function is also continuous at $x=-1.$ As the function is continuous at both points, it does not have any discontinuities.