One-sided limits
The instrument's function jumps at $x = 1$: from the left the outputs head toward 2, from the right toward 3. Does $\lim_{x \to 1} f(x)$ exist?
The two runs settle on different numbers. From the left: $f(0.9) = 1.9$, $f(0.99) = 1.99$ — the readout pins to 2. From the right: $f(1.1) = 2.9$, $f(1.01) = 2.99$ — it pins to 3. One input target, two output targets, and the marker never comes near 2.5.
Each run measured a one-sided limit. Write $x \to 1^-$ for the left approach and $x \to 1^+$ for the right. Then $\lim_{x \to 1^-} f(x) = 2$ and $\lim_{x \to 1^+} f(x) = 3$. Both exist; each is a perfectly good number.
The two-sided limit $\lim_{x \to 1} f(x)$ demands one number that every approach agrees on. Here the sides disagree, so it does not exist. It is not 2.5: averaging invents a value the function never approaches — no probe position produced an output near 2.5. The rule:
In the instrument, $\lim_{x \to 1^+} f(x) = 3$. What exactly does this claim?
Still with the instrument's function ($f(x) = x + 1$ for $x \lt 1$, $f(x) = 4 - x$ for $x \ge 1$): what are $\lim_{x \to 1^-} f(x)$ and $f(1)$?
New function, not from this lesson: $g(x) = \dfrac{|x|}{x}$. What is $\lim_{x \to 0} g(x)$?
A two-sided limit is an agreement. Check $x \to a^-$ and $x \to a^+$ separately: $\lim_{x \to a} f(x)$ exists exactly when both one-sided limits exist and are equal, and then it equals that shared number. When the sides disagree, the limit does not exist — never average.