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One-sided limits of the floor function at x = 3

Let f(x)=⌊x⌋f(x) = \lfloor x \rfloor, the greatest integer function (floor function), which rounds xx down to the nearest integer. For example, ⌊2.7⌋=2\lfloor 2.7 \rfloor = 2 and ⌊3.0⌋=3\lfloor 3.0 \rfloor = 3.

Consider the graph of f(x)=⌊x⌋f(x) = \lfloor x \rfloor near x=3x = 3.

One-sided limits of the floor function at x = 3 graph

Using the graph (or by evaluating f(x)f(x) at values like 2.9,2.99,2.9992.9, 2.99, 2.999 and 3.1,3.01,3.0013.1, 3.01, 3.001 as scratch work), determine lim⁡x→3−f(x)\lim_{x \to 3^-} f(x) and lim⁡x→3+f(x)\lim_{x \to 3^+} f(x). Based on these two one-sided limits, does lim⁡x→3f(x)\lim_{x \to 3} f(x) exist? If it exists, state its value; if not, explain why not.

Hint

Think about what integer ⌊x⌋\lfloor x \rfloor equals for xx values slightly less than 3 (like 2.9, 2.99) versus slightly greater than 3 (like 3.01, 3.1). Compare the two one-sided results before concluding about the two-sided limit.

Step-by-step solution

For xx slightly less than 3, such as x=2.9x = 2.9 or 2.992.99, the floor function gives ⌊x⌋=2\lfloor x \rfloor = 2 in every case, since these values have not yet reached 3. So as x→3−x \to 3^-, f(x)f(x) stays fixed at 22, giving lim⁡x→3−f(x)=2\lim_{x \to 3^-} f(x) = 2.

For xx slightly greater than 3, such as x=3.01x = 3.01 or 3.13.1, the floor function gives ⌊x⌋=3\lfloor x \rfloor = 3. So as x→3+x \to 3^+, f(x)f(x) stays fixed at 33, giving lim⁡x→3+f(x)=3\lim_{x \to 3^+} f(x) = 3. Since the left-hand limit (22) and right-hand limit (33) are not equal, the two-sided limit lim⁡x→3f(x)\lim_{x \to 3} f(x) does not exist.

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