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One-sided limits of the ceiling function at x = -2

The ceiling function, denoted ⌈x⌉\lceil x \rceil, rounds xx up to the nearest integer. For example, ⌈1.3⌉=2\lceil 1.3 \rceil = 2, ⌈2⌉=2\lceil 2 \rceil = 2, and ⌈−0.7⌉=0\lceil -0.7 \rceil = 0. Let f(x)=⌈x⌉f(x) = \lceil x \rceil.

Consider f(x)=⌈x⌉f(x) = \lceil x \rceil near x=−2x = -2.

One-sided limits of the ceiling function at x = -2 graph

Using the graph (or by testing values like x=−2.1,−2.01,−1.99,−1.9x = -2.1, -2.01, -1.99, -1.9), find lim⁡x→−2−f(x)\displaystyle\lim_{x \to -2^-} f(x) and lim⁡x→−2+f(x)\displaystyle\lim_{x \to -2^+} f(x). Based on these, does lim⁡x→−2f(x)\displaystyle\lim_{x \to -2} f(x) exist? If it exists, give its value; if not, explain why not.

Hint

Plug in values slightly less than −2-2 (like −2.01-2.01) to evaluate the left-hand limit, and values slightly greater than −2-2 (like −1.99-1.99) to evaluate the right-hand limit. Remember that the ceiling function rounds up to the nearest integer, and that the value of ff exactly at x=−2x=-2 does not matter for the limit.

Step-by-step solution

For xx slightly less than −2-2, such as x=−2.1x = -2.1 or x=−2.01x = -2.01, the ceiling function rounds up to the next integer, which is −2-2. So as x→−2−x \to -2^-, f(x)f(x) stays at −2-2, giving lim⁡x→−2−f(x)=−2\lim_{x\to -2^-} f(x) = -2.

For xx slightly greater than −2-2, such as x=−1.99x = -1.99 or x=−1.9x = -1.9, the ceiling rounds up to −1-1. So as x→−2+x \to -2^+, f(x)f(x) stays at −1-1, giving lim⁡x→−2+f(x)=−1\lim_{x\to -2^+} f(x) = -1. Since the left-hand limit (−2-2) and right-hand limit (−1-1) are not equal, the two-sided limit lim⁡x→−2f(x)\lim_{x\to -2} f(x) does not exist.

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