## Calculus 10th Edition

Published by Brooks Cole

# Chapter 1 - Limits and Their Properties - 1.4 Exercises: 53

#### Answer

The function has a jump discontinuity at $x=2.$ This discontinuity is irremovable.

#### Work Step by Step

For this piece-wise function, the only possible point where a discontinuity could occur is $x=2:$ $\lim\limits_{x\to2^+}f(x)=3-(2^+)=1; \lim\limits_{x\to2^-}f(x)=\frac{1}{2}(2^-)+1=2.$ Since $\lim\limits_{x\to2^-}f(x)\ne\lim\limits_{x\to2^+}f(x)\to\lim\limits_{x\to2}f(x)$ does not exist. Since the limit does not exist, the function is not continuous at the point $x=2.$

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