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Existence of a limit at a jump point of a ceiling function

Let a(x)=⌈2x⌉a(x) = \lceil 2x \rceil, the ceiling function applied to 2x2x (the ceiling function rounds a number up to the nearest integer).

Consider the function a(x)=⌈2x⌉.a(x) = \lceil 2x \rceil. Examine the behavior of a(x)a(x) for xx-values close to 1.51.5, both slightly less than and slightly greater than 1.51.5.

Existence of a limit at a jump point of a ceiling function graph

Does lim⁡x→1.5a(x)\lim_{x \to 1.5} a(x) exist? Justify your answer by finding lim⁡x→1.5−a(x)\lim_{x \to 1.5^-} a(x) and lim⁡x→1.5+a(x)\lim_{x \to 1.5^+} a(x), and state your final conclusion.

Hint

Pick xx-values just below 1.5 (like 1.4, 1.49, 1.499) and just above 1.5 (like 1.6, 1.51, 1.501), and compute ⌈2x⌉\lceil 2x \rceil for each to see what value a(x)a(x) approaches from each side.

Step-by-step solution

For xx slightly less than 1.51.5, say x=1.49x = 1.49, we get 2x=2.982x = 2.98, so ⌈2x⌉=3\lceil 2x \rceil = 3. As xx approaches 1.51.5 from the left, 2x2x approaches 33 from below, and the ceiling of any number strictly between 22 and 33 is 33. So lim⁡x→1.5−a(x)=3\lim_{x \to 1.5^-} a(x) = 3.

For xx slightly greater than 1.51.5, say x=1.51x = 1.51, we get 2x=3.022x = 3.02, so ⌈2x⌉=4\lceil 2x \rceil = 4. As xx approaches 1.51.5 from the right, 2x2x approaches 33 from above, and the ceiling of any number strictly between 33 and 44 is 44. So lim⁡x→1.5+a(x)=4\lim_{x \to 1.5^+} a(x) = 4. Since the left-hand limit (33) and right-hand limit (44) are not equal, the two-sided limit lim⁡x→1.5a(x)\lim_{x \to 1.5} a(x) does not exist.

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