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Classifying the discontinuity at x = 3

Consider the piecewise function y(x)={2x+1x<37x=3x+5x>3y(x) = \begin{cases} 2x+1 & x < 3 \\ 7 & x = 3 \\ x+5 & x > 3 \end{cases}

Determine whether y(x)y(x) is continuous at x=3x=3. If it is not continuous, classify the discontinuity as removable, jump, or infinite, and briefly justify your classification using the relevant one-sided limits and the value y(3)y(3).

Hint

Compute the left-hand limit using 2x+12x+1 and the right-hand limit using x+5x+5 as x→3x \to 3, then compare these to each other and to y(3)=7y(3) = 7. Recall that a jump discontinuity occurs when both one-sided limits exist but are unequal.

Step-by-step solution

First find the one-sided limits at x=3x=3. From the left, using 2x+12x+1: lim⁡x→3−y(x)=2(3)+1=7\lim_{x\to 3^-} y(x) = 2(3)+1 = 7. From the right, using x+5x+5: lim⁡x→3+y(x)=3+5=8\lim_{x\to 3^+} y(x) = 3+5 = 8.

Since the left-hand limit is 77 and the right-hand limit is 88, these two values are unequal, so lim⁡x→3y(x)\lim_{x\to 3} y(x) does not exist even though both one-sided limits exist and are finite. Because both one-sided limits are finite but different, this is a jump discontinuity at x=3x=3 (the value y(3)=7y(3)=7 is irrelevant to this classification, since the limit itself fails to exist).

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