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Checking the three-part continuity test at x = 1

Consider the piecewise function q(x)={x2−1x≤12x−2x>1q(x) = \begin{cases} x^2 - 1 & x \le 1 \\ 2x - 2 & x > 1 \end{cases}

Use the definition of continuity to determine whether q(x)q(x) is continuous at x=1x = 1. Specifically, find q(1)q(1), find lim⁡x→1q(x)\lim_{x \to 1} q(x) (by computing both one-sided limits), and state whether these two values are equal. Based on this, is qq continuous at x=1x = 1? If not, which of the three continuity conditions fails, and how would you classify the discontinuity (removable, jump, or infinite)?

Hint

Since qq is defined by two different formulas depending on whether x≤1x \le 1 or x>1x > 1, compute the left-hand limit using x2−1x^2-1 and the right-hand limit using 2x−22x-2, then compare both to q(1)q(1).

Step-by-step solution

Since 1≤11 \le 1, we use the first piece to get q(1)=12−1=0q(1) = 1^2 - 1 = 0. For the left-hand limit, as x→1−x \to 1^-, we also use x2−1x^2-1, giving lim⁡x→1−q(x)=12−1=0\lim_{x\to 1^-} q(x) = 1^2 - 1 = 0. For the right-hand limit, as x→1+x \to 1^+, we use 2x−22x-2, giving lim⁡x→1+q(x)=2(1)−2=0\lim_{x\to 1^+} q(x) = 2(1)-2 = 0.

Both one-sided limits equal 00, so lim⁡x→1q(x)=0\lim_{x\to 1} q(x) = 0 exists. Comparing this to q(1)=0q(1) = 0, the two values match. Since all three conditions (q(1)q(1) defined, limit exists, limit equals q(1)q(1)) hold, qq is continuous at x=1x=1, and there is no discontinuity to classify.

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