Q.Discuss the continuity of $f(x) = \dfrac{1}{x - 3}$, and state the interval(s) on which it is continuous.

This is a rational function, continuous wherever its denominator is non-zero; find where the denominator vanishes.

$f(x) = \dfrac{1}{x-3}$ is a rational function (a ratio of the constant polynomial $1$ to the polynomial $x - 3$). By the standard result of §4, a rational function is continuous at every point where its denominator is non-zero.

Locate the denominator's zero. Set $x - 3 = 0$, giving $x = 3$. This is the only point where the denominator vanishes.

Continuity away from $x = 3$. For every $x \neq 3$, the denominator is non-zero, so $f$ is continuous there. Thus $f$ is continuous at every point of its domain $\{x : x \neq 3\}$.

Behaviour at $x = 3$. At $x = 3$ the function is not even defined, so condition 1 fails and $f$ is discontinuous there. Examining the one-sided limits: as $x \to 3^+$, $x - 3 \to 0^+$ so $\dfrac{1}{x-3} \to +\infty$; as $x \to 3^-$, $x - 3 \to 0^-$ so $\dfrac{1}{x-3} \to -\infty$. Since a one-sided limit is infinite, this is an infinite discontinuity (a vertical asymptote at $x = 3$). …

A common slip is to say $f$ is "discontinuous everywhere" or "not a continuous function" because of the break at $x=3$. In fact $f$ is a continuous function on its domain — the point $x=3$ is not in the domain at all. Continuity is only ever assessed at points wh …