Define $f(x) = \dfrac{x^2 - 1}{x - 1}$ for $x \neq 1$ and $f(1) = 2$. Is $f$ continuous at $x = 1$?

**Definition.** $f$ is continuous at $x = 1$ iff $\displaystyle\lim_{x \to 1} f(x) = f(1)$. **Compute the limit.** For $x \neq 1$, $\dfrac{x^2 - 1}{x - 1} = x + 1$, so $\displaystyle\lim_{x \to 1} f(x) = 2$. **Compare to $f(1) = 2$.** They match, so $f$ is continuous at $x = 1$. **Common mistake.** Option B confuses *evaluation* of the formula at $x = 1$ (which is $0/0$, undefined) with the *limit* — but $f(1)$ is defined separately by the piecewise rule, and what matters for continuity is the limit, not the formula's raw value.