Q.Examine the continuity of the modulus function $f(x) = |x - 2|$ at $x = 2$.

Write the modulus as a piecewise rule around $x=2$ and check the two one-sided limits against $f(2)$.

The modulus $|x-2|$ splits into two rules at $x = 2$:

Value at the point. $f(2) = |2 - 2| = 0$.

Right-hand limit. As $x \to 2^+$, the rule $x - 2$ applies: $\displaystyle\lim_{x\to 2^+} f(x) = 2 - 2 = 0$.

Left-hand limit. As $x \to 2^-$, the rule $2 - x$ applies: $\displaystyle\lim_{x\to 2^-} f(x) = 2 - 2 = 0$.

Compare. Both one-sided limits equal $0$, so the two-sided limit exists and equals $0$; and $f(2) = 0$ matches it. All three conditions hold, so $f$ is continuous at $x = 2$. …

Shortcut via §4–§5: $|x-2|$ is a modulus (continuous everywhere) applied to a polynomial (continuous everywhere), hence continuous at every real num …

A common mistake is to think the "corner" in the graph of $|x-2|$ at $x=2$ signals a discontinuity. A sharp corner affects differentiability, not continuity — the graph is still drawn without lifting the pen. The function is perfectly continuo …