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Continuity of the floor function at x = 3

Let s(x)=⌊x⌋s(x) = \lfloor x \rfloor, the floor function (also called the greatest integer function), which rounds xx down to the nearest integer. For example, ⌊2.7⌋=2\lfloor 2.7 \rfloor = 2 and ⌊3⌋=3\lfloor 3 \rfloor = 3.

Consider s(x)=⌊x⌋s(x) = \lfloor x \rfloor at a=3a = 3.

Continuity of the floor function at x = 3 graph

Using the three-part 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 conditions fails, evaluate lim⁡x→3−s(x)\lim_{x \to 3^-} s(x) and lim⁡x→3+s(x)\lim_{x \to 3^+} s(x), and classify the type of discontinuity (removable, jump, or infinite).

Hint

Compute s(3)s(3) directly from the definition, then examine what happens to s(x)s(x) as xx approaches 3 from values slightly less than 3 versus values slightly greater than 3.

Step-by-step solution

First check condition 1: s(3)=⌊3⌋=3s(3) = \lfloor 3 \rfloor = 3, so f(a)f(a) is defined. Next examine the one-sided limits. For xx slightly less than 3 (like 2.9,2.992.9, 2.99), ⌊x⌋=2\lfloor x \rfloor = 2, so lim⁡x→3−s(x)=2\lim_{x \to 3^-} s(x) = 2. For xx slightly greater than 3 (like 3.01,3.13.01, 3.1), ⌊x⌋=3\lfloor x \rfloor = 3, so lim⁡x→3+s(x)=3\lim_{x \to 3^+} s(x) = 3.

Since the left-hand limit (2) and right-hand limit (3) are different, lim⁡x→3s(x)\lim_{x\to 3} s(x) does not exist, so condition 2 of the continuity definition fails. Because both one-sided limits are finite but unequal, this is a jump discontinuity at x=3x = 3, and therefore s(x)s(x) is not continuous there.

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