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Prove that \(f(x) = \begin{cases} 2-x, & \quad \text{when x <2;} \text{}\\ 2+x, & \quad \text{whenx≥2} \end{cases}\) is discontinuous at x=2 |
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Answer» LHL: = \(\lim\limits_{x \to2^-} \)f(x) = \(\lim\limits_{x \to2^-} \) 2 + x = 4 RHL: = \(\lim\limits_{x \to2^+} \)f(x) = \(\lim\limits_{x \to2^+} \) 2 - x = 0 = \(\lim\limits_{x \to2^-} \)f(x) ≠ \(\lim\limits_{x \to2^+} \) f(x) f(x) is discontinuous at x=2 |
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