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Classify the discontinuity of a piecewise function at a breakpoint

Consider the piecewise function h(x)={3x+1x<2x2x≥2h(x) = \begin{cases} 3x + 1 & x < 2 \\ x^2 & x \ge 2 \end{cases}

Determine whether h(x)h(x) is continuous at x=2x = 2. If it is not continuous, classify the discontinuity at x=2x = 2 as removable, jump, or infinite, and justify your classification using the one-sided limits of hh at x=2x = 2.

Hint

Compute lim⁡x→2−h(x)\lim_{x \to 2^-} h(x) using 3x+13x+1 and lim⁡x→2+h(x)\lim_{x \to 2^+} h(x) using x2x^2. Compare these two one-sided limits to each other, and also compare their common value (if any) to h(2)h(2).

Step-by-step solution

First find the one-sided limits at x=2x=2. From the left, h(x)=3x+1h(x) = 3x+1, so lim⁡x→2−h(x)=3(2)+1=7\lim_{x \to 2^-} h(x) = 3(2)+1 = 7. From the right, h(x)=x2h(x) = x^2, so lim⁡x→2+h(x)=(2)2=4\lim_{x \to 2^+} h(x) = (2)^2 = 4.

Since the left-hand limit (77) and right-hand limit (44) are both finite but not equal to each other, lim⁡x→2h(x)\lim_{x\to 2} h(x) does not exist, and the graph has a break where the two pieces do not meet. Because both one-sided limits are finite (no infinite behavior) but disagree, this is a jump discontinuity at x=2x=2.

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