Finding Constants that Make a Function Continuous

A very common problem type gives a function containing one or more unknown constants and asks for the value(s) that make the function continuous at a stated point (or across its whole domain). The method is a direct application of the three-condition test: set the relevant limit equal to the value the function is required to take, and solve for the unknown.

Case A — a single point, one unknown (removable-type). The function is defined by a rule for $x \neq a$ and by an unknown value $k$ at $x = a$. Continuity at $a$ requires $\displaystyle\lim_{x \to a} f(x) = f(a) = k$. So:

1. Compute $\displaystyle\lim_{x \to a} f(x)$ (usually by cancelling a common factor, or by rationalising a surd, exactly as in the Limits chapter).
2. Set $k$ equal to that limit. That single value of $k$ makes $f$ continuous.

Case B — a joining point of a piecewise function, one unknown. The function switches rule at $x = a$, with an unknown constant in one of the pieces. Continuity requires the left-hand limit, the right-hand limit and $f(a)$ to be equal. Set the two relevant expressions equal at $x = a$ and solve for the unknown.

Case C — two joining points, two unknowns. The function has three pieces joined at two points, with two unknown constants (say $a$ and $b$). Apply the continuity condition at each joining point; this gives two linear equations in the two unknowns, which are then solved simultaneously.

Illustration (Case A). Let $f(x) = \dfrac{x^2 - 1}{x - 1}$ for $x \neq 1$, and $f(1) = k$. For $x \neq 1$, $\dfrac{x^2-1}{x-1} = \dfrac{(x-1)(x+1)}{x-1} = x + 1$, so $\displaystyle\lim_{x \to 1} f(x) = 1 + 1 = 2$. Continuity at $x = 1$ requires $k = 2$. Any other value of $k$ leaves a removable discontinuity at $x = 1$.