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Classifying the discontinuity of the floor function at x = 3

Let s(x)=⌊x⌋s(x) = \lfloor x \rfloor, the greatest integer function (also called the floor function), which outputs the greatest integer less than or equal to xx.

Consider s(x)=⌊x⌋s(x) = \lfloor x \rfloor at x=3x = 3. Recall that ⌊x⌋\lfloor x \rfloor equals 22 for all xx in [2,3)[2,3) and equals 33 for all xx in [3,4)[3,4).

Classifying the discontinuity of the floor function at x = 3 graph

Using the definition of continuity, determine whether s(x)s(x) is continuous at x=3x = 3. If it is not continuous, state which of the three continuity conditions fails, and classify the discontinuity as removable, jump, or infinite.

Hint

Compute lim⁡x→3−s(x)\lim_{x \to 3^-} s(x) and lim⁡x→3+s(x)\lim_{x \to 3^+} s(x) separately, along with s(3)s(3), then check the three continuity conditions in order.

Step-by-step solution

First check the pieces: for xx slightly less than 3, s(x)=⌊x⌋=2s(x) = \lfloor x \rfloor = 2, so lim⁡x→3−s(x)=2\lim_{x \to 3^-} s(x) = 2. For xx slightly greater than or equal to 3, s(x)=3s(x) = 3, so lim⁡x→3+s(x)=3\lim_{x \to 3^+} s(x) = 3. Also, s(3)=⌊3⌋=3s(3) = \lfloor 3 \rfloor = 3, so f(a)f(a) is defined.

Since the one-sided limits 22 and 33 are not equal, lim⁡x→3s(x)\lim_{x \to 3} s(x) does not exist — this is the continuity condition that fails (the second condition). Because both one-sided limits exist (they are finite) but disagree, this is a jump discontinuity, not removable (since no single limit value exists to compare to f(3)f(3)) and not infinite (since neither one-sided limit is infinite).

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