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Existence of a two-sided limit at a jump discontinuity

Let p(x)=⌊x⌋p(x) = \lfloor x \rfloor be the floor function, which rounds xx down to the nearest integer. For example, ⌊2.3⌋=2\lfloor 2.3 \rfloor = 2, ⌊2.9⌋=2\lfloor 2.9 \rfloor = 2, and ⌊3⌋=3\lfloor 3 \rfloor = 3.

Consider the function p(x)=⌊x⌋p(x) = \lfloor x \rfloor near x=3x = 3.

Existence of a two-sided limit at a jump discontinuity graph

Using the graph (or by testing values of xx just below and just above 3), determine lim⁡x→3−p(x)\lim_{x \to 3^-} p(x) and lim⁡x→3+p(x)\lim_{x \to 3^+} p(x). Based on these, does lim⁡x→3p(x)\lim_{x \to 3} p(x) exist? If it exists, state its value; if not, explain why not.

Hint

Evaluate p(x)p(x) for values slightly less than 3 (like 2.9, 2.99) and slightly greater than 3 (like 3.01, 3.1) separately, then compare the two trends.

Step-by-step solution

For xx slightly less than 3, such as x=2.9x = 2.9 or x=2.99x = 2.99, we have ⌊x⌋=2\lfloor x \rfloor = 2 since these values have not yet reached 3. So as x→3−x \to 3^-, p(x)p(x) stays at 2, giving lim⁡x→3−p(x)=2\lim_{x \to 3^-} p(x) = 2.

For xx slightly greater than 3, such as x=3.01x = 3.01 or x=3.1x = 3.1, we have ⌊x⌋=3\lfloor x \rfloor = 3. So as x→3+x \to 3^+, p(x)p(x) stays at 3, giving lim⁡x→3+p(x)=3\lim_{x \to 3^+} p(x) = 3. Since the left-hand limit (2) and right-hand limit (3) are not equal, the two-sided limit lim⁡x→3p(x)\lim_{x \to 3} p(x) does not exist.

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