## Calculus 10th Edition

$|x-3|\leq2\to-2\leq x-3\leq2\to1\leq x\leq5\to$ the possible x-values for discontinuities are $x=1$ and $x=5:$ $\lim\limits_{x\to1^-}f(x)=2; \lim\limits_{x\to1^+}f(x)=\csc{\dfrac{\pi (1^+)}{6}}=2.$ Since $\lim\limits_{x\to1^-}f(x)=\lim\limits_{x\to1^+}f(x)\to\lim\limits_{x\to1}f(x)=2=f(1)$ hence the function is continuous at $x=1.$ $\lim\limits_{x\to5^-}f(x)=\csc{\dfrac{\pi(5^-)}{6}}=2; \lim\limits_{x\to5^+}f(x)=2.$ Since $\lim\limits_{x\to5^-}f(x)=\lim\limits_{x\to5^+}f(x)\to\lim\limits_{x\to5}f(x)=2=f(5)$ hence the function is continuous at $x=5.$ As the function is continuous at both points, it does not have any discontinuities.